Math24.net
Calculus
Home
Calculus
Limits and Continuity
Differentiation
Applications of Derivative
Integration
Sequences and Series
Double Integrals
Triple Integrals
Line Integrals
Surface Integrals
Fourier Series
Differential Equations
1st Order Equations
2nd Order Equations
Nth Order Equations
Systems of Equations
Formulas and Tables
   Integration of Rational Functions
A rational function \(\large\frac{{P\left( x \right)}}{{Q\left( x \right)}}\normalsize,\) where \({P\left( x \right)}\) and \({Q\left( x \right)}\) are both polynomials, can be integrated in four steps:
  1. Reduce the fraction if it is improper (i.e. degree of \({P\left( x \right)}\) is greater than degree of \({Q\left( x \right)}\);

  2. Factor \({Q\left( x \right)}\) into linear and/or quadratic (irreducible) factors;

  3. Decompose the fraction into a sum of partial fractions;

  4. Calculate integrals of each partial fraction.

Consider the specified steps in more details.

Step 1. Reducing an improper fraction
If the fraction is improper (i.e. degree of \({P\left( x \right)}\) is greater than degree of \({Q\left( x \right)}\)), divide the numerator \({P\left( x \right)}\) by the denominator \({Q\left( x \right)}\) to obtain \[\frac{{P\left( x \right)}}{{Q\left( x \right)}} = F\left( x \right) + \frac{{R\left( x \right)}}{{Q\left( x \right)}},\] where \(\large\frac{{R\left( x \right)}}{{Q\left( x \right)}}\normalsize\) is a proper fraction.

Step 2. Factoring \({Q\left( x \right)}\) into linear and/or quadratic factors
Write the denominator \({Q\left( x \right)}\) as \[ {Q\left( x \right) } = {{\left( {x - a} \right)^\alpha } \cdots {\left( {x - b} \right)^\beta }{\left( {{x^2} + px + q} \right)^\mu } \cdots {\left( {{x^2} + rx + s} \right)^\nu },} \] where quadratic functions are irreducible, i.e. do not have real roots.

Step 3. Decomposing the rational fraction into a sum of partial fractions.
Write the function as follows: \[ {\frac{{R\left( x \right)}}{{Q\left( x \right)}} = \frac{A}{{{{\left( {x - a} \right)}^\alpha }}} + \frac{{{A_1}}}{{{{\left( {x - a} \right)}^{\alpha - 1}}}} + \ldots } {+ \frac{{{A_{\alpha - 1}}}}{{x - a}} + \ldots } {+ \frac{B}{{{{\left( {x - b} \right)}^\beta }}} + \frac{{{B_1}}}{{{{\left( {x - b} \right)}^{\beta - 1}}}} + \ldots } {+ \frac{{{B_{\beta - 1}}}}{{x - b}} } {+ \frac{{Kx + L}}{{{{\left( {{x^2} + px + q} \right)}^\mu }}} + \frac{{{K_1}x + {L_1}}}{{{{\left( {{x^2} + px + q} \right)}^{\mu - 1}}}} + \ldots } {+ \frac{{{K_{\mu - 1}}x + {L_{\mu - 1}}}}{{{x^2} + px + q}} + \ldots } {+ \frac{{Mx + N}}{{{{\left( {{x^2} + rx + s} \right)}^\nu }}} + \frac{{{M_1}x + {N_1}}}{{{{\left( {{x^2} + rx + s} \right)}^{\nu - 1}}}} + \ldots } {+ \frac{{{M_{\nu - 1}}x + {N_{\nu - 1}}}}{{{x^2} + rx + s}}.} \] The total number of undetermined coefficients \({A_i},\) \({B_i},\) \({K_i},\) \({L_i},\) \({M_i},\) \({N_i}, \ldots\) must be equal to the degree of the denominator \({Q\left( x \right)}.\)

Then equate the coefficients of equal powers of \(x\) by multiplying both sides of the latter expression by \({Q\left( x \right)}\) and write the system of linear equations in \({A_i},\) \({B_i},\) \({K_i},\) \({L_i},\) \({M_i},\) \({N_i}, \ldots\) The resulting system must always have a unique solution.

Step 4. Integrating partial fractions.
Use the following \(6\) formulas to evaluate integrals of partial fractions with linear and quadratic denominators: \[1.\;\;\int {\frac{A}{{x - a}}dx} = \ln \left| {x - a} \right|\] \[2.\;\;\int {\frac{A}{{{{\left( {x - a} \right)}^k}}}dx} = \frac{1}{{\left( {1 - k} \right){{\left( {x - a} \right)}^{k - 1}}}}\] For fractions with quadratic denominators, first complete the square: \[ {\int {\frac{{Ax + B}}{{{{\left( {{x^2} + px + q} \right)}^k}}}dx} } = {\int {\frac{{At + B'}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}dt} ,} \] where \(t = x + {\large\frac{p}{2}\normalsize},\) \({m^2} = {\large\frac{{4q - {p^2}}}{4}\normalsize},\) \(B' = B - {\large\frac{{Ap}}{2}\normalsize}.\) Then use the formulas: \[3.\;\;\int {\frac{{tdt}}{{{t^2} + {m^2}}}} = \frac{1}{2}\ln \left( {{t^2} + {m^2}} \right)\] \[ {4.\;\;\int {\frac{{tdt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} } = {\frac{1}{{2\left( {1 - k} \right){{\left( {{t^2} + {m^2}} \right)}^{k - 1}}}} } \] \[5.\;\;\int {\frac{{dt}}{{{t^2} + {m^2}}}} = \frac{1}{a}\arctan \frac{t}{m}\] The integral \(\large\int\normalsize {\large\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}\normalsize} \) can be calculated in \(k\) steps using the reduction formula: \[ {6.\;\;\int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} } = {\frac{t}{{2{m^2}\left( {k - 1} \right){{\left( {{t^2} + {m^2}} \right)}^{k - 1}}}} } {+ \frac{{2k - 3}}{{2{m^2}\left( {k - 1} \right)}}\int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^{k - 1}}}}} } \]
   Example 1
Evaluate the integral \({\large\int\normalsize} {{\large\frac{{2x + 3}}{{{x^2} - 9}}\normalsize} dx} .\)

Solution.
Decompose the integrand into partial functions: \[ {\frac{{2x + 3}}{{{x^2} - 9}} } = {\frac{{2x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} } = {\frac{A}{{x - 3}} + \frac{B}{{x + 3}}.} \] Equate coefficients: \[ {A\left( {x + 3} \right) + B\left( {x - 3} \right) = 2x + 3,}\;\; {\Rightarrow Ax + 3A + Bx - 3B = 2x + 3,}\;\; {\Rightarrow \left( {A + B} \right)x + 3A - 3B = 2x + 3.} \] Hence, \[ {\left\{ \begin{array}{l} A + B = 2\\ 3A - 3B = 3 \end{array} \right.,}\;\; {\Rightarrow \left\{ \begin{array}{l} A = \frac{3}{2}\\ B = \frac{1}{2} \end{array} \right..} \] Then \[ {\frac{{2x + 3}}{{{x^2} - 9}} } = {\frac{{\frac{3}{2}}}{{x - 3}} + \frac{{\frac{1}{2}}}{{x + 3}}.} \] The integral is equal to \[ {\int {\frac{{2x + 3}}{{{x^2} - 9}}dx} } = {\frac{3}{2}\int {\frac{{dx}}{{x - 3}}} + \frac{1}{2}\int {\frac{{dx}}{{x + 3}}} } = {\frac{3}{2}\ln \left| {x - 3} \right| + \frac{1}{2}\ln \left| {x + 3} \right| + C } = {\frac{1}{2}\ln \left| {{{\left( {x - 3} \right)}^2}\left( {x + 3} \right)} \right| + C.} \]
   Example 2
Evaluate \({\large\int\normalsize} {{\large\frac{{{x^2} - 2}}{{x + 1}}\normalsize} dx}.\)

Solution.
First we divide the numerator by the denominator, obtaining \[\frac{{{x^2} - 2}}{{x + 1}} = x - 1 - \frac{1}{{x + 1}}.\] Then \[ {\int {\frac{{{x^2} - 2}}{{x + 1}}dx} } = {\int {\left( {x - 1 - \frac{1}{{x + 1}}} \right)dx} } = {\int {xdx} - \int {dx} - \int {\frac{{dx}}{{x + 1}}} } = {\frac{{{x^2}}}{2} - x - \ln \left| {x + 1} \right| + C.} \]
   Example 3
Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{x^2} + 4x + 8}}\normalsize}.\)

Solution.
We can write: \[ {\int {\frac{{dx}}{{{x^2} + 4x + 8}}} } = {\int {\frac{{dx}}{{{x^2} + 4x + 4 + 4}}} } = {\int {\frac{{dx}}{{{{\left( {x + 2} \right)}^2} + 4}}} } = {\int {\frac{{d\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2} + {2^2}}}} } = {\frac{1}{2}\arctan \frac{{x + 2}}{2} + C.} \]
   Example 4
Evaluate the integral \({\large\int\normalsize} {\large\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}\normalsize} .\)

Solution.
Decompose the integrand into partial functions: \[ {\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} } = {\frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{C}{{x - 3}}.} \] Equate coefficients: \[ {A\left( {x - 2} \right)\left( {x - 3} \right) } {+ B\left( {x - 1} \right)\left( {x - 3} \right) } {+ C\left( {x - 1} \right)\left( {x - 2} \right) } = {{x^2},} \] \[ {A{x^2} - 2Ax - 3Ax } {+ 6A + B{x^2} - Bx - 3Bx } {+ 3B + C{x^2} - Cx - 2Cx + 2C } = {{x^2},} \] \[ {\left( {A + B + C} \right){x^2} } {- \left( {5A + 4B + 3C} \right)x } {+ 6A + 3B + 2C } = {{x^2}.} \] Hence, \[ {\left\{ \begin{array}{l} A + B + C = 1\\ 5A + 4B + 3C = 0\\ 6A + 3B + 2C = 0 \end{array} \right.,}\;\; {\Rightarrow \left\{ \begin{array}{l} A = \frac{1}{2}\\ B = - 4\\ C = \frac{9}{2} \end{array} \right..} \] Then \[ {\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} } = {\frac{{\frac{1}{2}}}{{x - 1}} - \frac{4}{{x - 2}} + \frac{{\frac{9}{2}}}{{x - 3}}.} \] The integral is equal to \[ {\int {\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}} } = {\frac{1}{2}\int {\frac{{dx}}{{x - 1}}} - 4\int {\frac{{dx}}{{x - 2}}} + \frac{9}{2}\int {\frac{{dx}}{{x - 3}}} } = {\frac{1}{2}\ln \left| {x - 1} \right| - 4\ln \left| {x - 2} \right| + \frac{9}{2}\ln \left| {x - 3} \right| + C.} \]
   Example 5
Evaluate \({\large\int\normalsize} {\large\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}\normalsize} .\)

Solution.
Decompose the integrand into the sum of two fractions: \[ {\frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} } = {\frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 1}}.} \] Equate coefficients: \[ {A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {x + 1} \right) = 1,}\;\; {\Rightarrow A{x^2} + A + B{x^2} + Cx + Bx + C = 1,}\;\; {\Rightarrow \left( {A + B} \right){x^2} + \left( {B + C} \right)x + A + C = 1.} \] Hence \[ {\left\{ \begin{array}{l} A + B = 0\\ B + C = 0\\ A + C = 1 \end{array} \right.,}\;\; {\Rightarrow \left\{ \begin{array}{l} A = \frac{1}{2}\\ B = - \frac{1}{2}\\ C = \frac{1}{2} \end{array} \right..} \] The integrand can be written as \[ {\frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} } = {\frac{{\frac{1}{2}}}{{x + 1}} + \frac{{ - \frac{1}{2}x + \frac{1}{2}}}{{{x^2} + 1}} } = {\frac{1}{{2\left( {x + 1} \right)}} - \frac{1}{2} \cdot \frac{x}{{{x^2} + 1}} + \frac{1}{2} \cdot \frac{1}{{{x^2} + 1}}.} \] The initial integral becomes \[ {\int {\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}} } = {\frac{1}{2}\int {\frac{{dx}}{{x + 1}}} - \frac{1}{2}\int {\frac{{xdx}}{{{x^2} + 1}}} + \frac{1}{2}\int {\frac{{dx}}{{{x^2} + 1}}} } = {\frac{1}{2}\ln \left| {x + 1} \right| - \frac{1}{4}\int {\frac{{d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} + \frac{1}{2}\arctan x } = {\frac{1}{2}\ln \left| {x + 1} \right| - \frac{1}{4}\ln \left( {{x^2} + 1} \right) + \frac{1}{2}\arctan x + C.} \]
   Example 6
Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{x^3} + 1}}\normalsize} .\)

Solution.
We can factor the denominator in the integrand: \[{x^3} + 1 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right).\] Decompose the integrand into partial functions: \[ {\frac{1}{{{x^3} + 1}} } = {\frac{1}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} } = {\frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} - x + 1}}.} \] Equate coefficients: \[ {A\left( {{x^2} - x + 1} \right) + \left( {Bx + C} \right)\left( {x + 1} \right) = 1,}\;\; {\Rightarrow A{x^2} - Ax + A + B{x^2} + Cx + Bx + C = 1,}\;\; {\Rightarrow \left( {A + B} \right){x^2} + \left( { - A + B + C} \right)x + A + C = 1.} \] Hence, \[ {\left\{ \begin{array}{l} A + B = 0\\ - A + B + C = 0\\ A + C = 1 \end{array} \right.,}\;\; {\Rightarrow \left\{ \begin{array}{l} A = \frac{1}{3}\\ B = - \frac{1}{3}\\ C = \frac{2}{3} \end{array} \right..} \] Then \[ {\frac{1}{{{x^3} + 1}} } = {\frac{{\frac{1}{3}}}{{x + 1}} + \frac{{ - \frac{1}{3}x + \frac{2}{3}}}{{{x^2} - x + 1}} } = {\frac{1}{{3\left( {x + 1} \right)}} - \frac{1}{3} \cdot \frac{{x - 2}}{{{x^2} - x + 1}}.} \] Now we can calculate the initial integral: \[ {\int {\frac{{dx}}{{{x^3} + 1}}} } = {\frac{1}{3}\int {\frac{{dx}}{{x + 1}}} - \frac{1}{3}\int {\frac{{x - 2}}{{{x^2} - x + 1}}dx} } = {\frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{3}\int {\frac{{x - \frac{1}{2} - \frac{3}{2}}}{{{x^2} - x + 1}}dx} } = {\frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{3}\int {\frac{{x - \frac{1}{2}}}{{{x^2} - x + 1}}dx} } {+ \frac{1}{2}\int {\frac{{dx}}{{{x^2} - x + 1}}} } = {\frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{6}\int {\frac{{\left( {2x - 1} \right)dx}}{{{x^2} - x + 1}}} } {+ \frac{1}{2}\int {\frac{{dx}}{{{{\left( {x - \frac{1}{2}} \right)}^2} + \frac{3}{4}}}} } = {\frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{6}\int {\frac{{d\left( {{x^2} - x + 1} \right)}}{{{x^2} - x + 1}}} } {+ \frac{1}{2}\int {\frac{{d\left( {x - \frac{1}{2}} \right)}}{{{{\left( {x - \frac{1}{2}} \right)}^2} } {+ {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} } = {\frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{6}\ln \left( {{x^2} - x + 1} \right) } {+ \frac{1}{{\sqrt 3 }}\arctan \frac{{2x - 1}}{{\sqrt 3 }} + C.} \]
   Example 7
Calculate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{x^4} + 1}}\normalsize}. \)

Solution.
Rewrite the denominator in the integrand as follows: \[ {{x^4} + 1 } = {{x^4} + 2{x^2} - 2{x^2} + 1 } = {\left( {{x^4} + 2{x^2} + 1} \right) - 2{x^2} } = {{\left( {{x^2} + 1} \right)^2} - 2{x^2} } = {\left( {{x^2} + 1 - \sqrt 2 x} \right)\left( {{x^2} + 1 + \sqrt 2 x} \right).} \] The factors in the denominator are irreducible quadratic factors since they have no real roots. Then \[ {\frac{1}{{{x^4} + 1}} } = {\frac{1}{{\left( {{x^2} + 1 - \sqrt 2 x} \right)\left( {{x^2} + 1 + \sqrt 2 x} \right)}} } = {\frac{{Ax + B}}{{{x^2} - \sqrt 2 x + 1}} + \frac{{Cx + D}}{{{x^2} + \sqrt 2 x + 1}}.} \] Equate coefficients: \[ {\left( {Ax + B} \right)\left( {{x^2} + \sqrt 2 x + 1} \right) } {+ \left( {Cx + D} \right)\left( {{x^2} - \sqrt 2 x + 1} \right) = 1,} \] \[ {A{x^3} + B{x^2} + \sqrt 2 A{x^2} } {+ \sqrt 2 Bx + Ax + B + C{x^3} } {+ D{x^2} - \sqrt 2 C{x^2} } {- \sqrt 2 Dx + Cx + D = 1,} \] \[ {\left( {A + C} \right){x^3} } {+ \left( {B + \sqrt 2 A + D - \sqrt 2 C} \right){x^2} } {+ \left( {\sqrt 2 B + A - \sqrt 2 D + C} \right)x } {+ B + D = 1.} \] This yields \[ {\left\{ \begin{array}{l} A + C = 0\\ B + \sqrt 2 A + D - \sqrt 2 C = 0\\ \sqrt 2 B + A - \sqrt 2 D + C = 0\\ B + D = 1 \end{array} \right.,}\;\; {\Rightarrow \left\{ \begin{array}{l} A = - \frac{1}{{2\sqrt 2 }}\\ B = \frac{1}{2}\sqrt 2 \\ C = \frac{1}{{2\sqrt 2 }}\\ D = \frac{1}{2} \end{array} \right..} \] Hence, \[ {\frac{1}{{{x^4} + 1}} } = {\frac{{ - \frac{1}{{2\sqrt 2 }}x + \frac{1}{2}}}{{{x^2} - \sqrt 2 x + 1}} + \frac{{\frac{1}{{2\sqrt 2 }}x + \frac{1}{2}}}{{{x^2} + \sqrt 2 x + 1}} } = { - \frac{1}{{2\sqrt 2 }}\frac{{x - \sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}} + \frac{1}{{2\sqrt 2 }}\frac{{x + \sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}}.} \] Integrating term by term, we obtain the answer: \[ {\int {\frac{{dx}}{{{x^4} + 1}}} } = {{ - \frac{1}{{2\sqrt 2 }}\int {\frac{{x - \sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}}dx} } + {\frac{1}{{2\sqrt 2 }}\int {\frac{{x + \sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}}dx} }} = {{ - \frac{1}{{4\sqrt 2 }}\int {\frac{{2x - \sqrt 2 - \sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}}dx} } + {\frac{1}{{4\sqrt 2 }}\int {\frac{{2x + \sqrt 2 + \sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}}dx} }} = {{ - \frac{1}{{4\sqrt 2 }}\int {\frac{{2x - \sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}}dx} } + {\frac{1}{{4\sqrt 2 }}\int {\frac{{\sqrt 2 }}{{{x^2} - \sqrt 2 x + 1}}dx} } + {\frac{1}{{4\sqrt 2 }}\int {\frac{{2x + \sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}}dx} } + {\frac{1}{{4\sqrt 2 }}\int {\frac{{\sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}}dx} }} = {{ - \frac{1}{{4\sqrt 2 }}\int {\frac{{d\left( {{x^2} - \sqrt 2 x + 1} \right)}}{{{x^2} - \sqrt 2 x + 1}}} } + {\frac{1}{4}\int {\frac{{dx}}{{{x^2} - \sqrt 2 x + 1}}} } + {\frac{1}{{4\sqrt 2 }}\int {\frac{{d\left( {{x^2} + \sqrt 2 x + 1} \right)}}{{{x^2} + \sqrt 2 x + 1}}} } + {\frac{1}{4}\int {\frac{{dx}}{{{x^2} + \sqrt 2 x + 1}}} }} = {{ - \frac{1}{{4\sqrt 2 }}\ln \left( {{x^2} - \sqrt 2 x + 1} \right) } + {\frac{1}{4}\int {\frac{{dx}}{{{{\left( {x - \frac{1}{{\sqrt 2 }}} \right)}^2} + \frac{1}{2}}}} } + {\frac{1}{{4\sqrt 2 }}\ln \left( {{x^2} + \sqrt 2 x + 1} \right) } + {\frac{1}{4}\int {\frac{{dx}}{{{{\left( {x + \frac{1}{{\sqrt 2 }}} \right)}^2} + \frac{1}{2}}}} }} = {{\frac{1}{{4\sqrt 2 }}\ln \left( {\frac{{{x^2} + \sqrt 2 x + 1}}{{{x^2} - \sqrt 2 x + 1}}} \right) } + {\frac{{\sqrt 2 }}{4}\arctan \frac{{x - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }}}} } + {\frac{{\sqrt 2 }}{4}\arctan \frac{{x + \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }}}} + C }} = {{\frac{1}{{4\sqrt 2 }}\ln \left( {\frac{{{x^2} + \sqrt 2 x + 1}}{{{x^2} - \sqrt 2 x + 1}}} \right) } + {\frac{1}{{2\sqrt 2 }}\left[ {\arctan \left( {\sqrt 2 x - 1} \right) + \arctan \left( {\sqrt 2 x + 1} \right)} \right] + C.}} \] We can simplify this answer. Let \[\alpha = \arctan \left( {\sqrt 2 x - 1} \right) + \arctan \left( {\sqrt 2 x + 1} \right).\] Then \[\require{cancel} {\tan \alpha = \tan \left[ {\arctan \left( {\sqrt 2 x - 1} \right) + \arctan \left( {\sqrt 2 x + 1} \right)} \right] } = {\frac{{\tan \left( {\arctan \left( {\sqrt 2 x - 1} \right)} \right) + \tan \left( {\arctan \left( {\sqrt 2 x + 1} \right)} \right)}}{{1 - \tan \left( {\arctan \left( {\sqrt 2 x - 1} \right)} \right) \cdot \tan \left( {\arctan \left( {\sqrt 2 x + 1} \right)} \right)}} } = {\frac{{\sqrt 2 x - \cancel{1} + \sqrt 2 x + \cancel{1}}}{{1 - \left( {\sqrt 2 x - 1} \right)\left( {\sqrt 2 x + 1} \right)}} } = {\frac{{2\sqrt 2 x}}{{1 - \left( {2{x^2} - 1} \right)}} } = {\frac{{\sqrt 2 x}}{{1 - {x^2}}}.} \] Hence, \(\alpha = \arctan {\large\frac{{\sqrt 2 x}}{{1 - {x^2}}}\normalsize}.\) The complete answer is \[ {I = \frac{1}{{4\sqrt 2 }}\ln \left( {\frac{{{x^2} + \sqrt 2 x + 1}}{{{x^2} - \sqrt 2 x + 1}}} \right) } + {\frac{1}{{2\sqrt 2 }}\arctan \frac{{\sqrt 2 x}}{{1 - {x^2}}} + C.} \]
   Example 8
Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{x^4} - 1}}\normalsize} .\)

Solution.
We can factor the denominator in the integrand: \[ {{x^4} - 1 } = {\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) } = {\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right).} \] Decompose the integrand into partial functions: \[ {\frac{1}{{{x^4} - 1}} } = {\frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} } = {\frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{{Cx + D}}{{{x^2} + 1}}.} \] Equate coefficients: \[ {A\left( {{x^2} + 1} \right)\left( {x + 1} \right) } {+ B\left( {{x^2} + 1} \right)\left( {x - 1} \right) } {+ \left( {Cx + D} \right)\left( {{x^2} - 1} \right) = 1,} \] \[ {A{x^3} + Ax + A{x^2} } {+ A + B{x^3} - B{x^2} } {+ Bx - B + C{x^3} } {+ D{x^2} - Cx - D = 1,} \] \[ {\left( {A + B + C} \right){x^3} } {+ \left( {A - B + D} \right){x^2}} {+ \left( {A + B - C} \right)x } {+ A - B - D = 1.} \] Hence, \[ {\left\{ \begin{array}{l} A + B + C = 0\\ A - B + D = 0\\ A + B - C = 0\\ A - B - D = 1 \end{array} \right.,}\;\; {\Rightarrow \left\{ \begin{array}{l} A = \frac{1}{4}\\ B = - \frac{1}{4}\\ C = 0\\ D = - \frac{1}{2} \end{array} \right..} \] Thus, the integrand becomes \[ {\frac{1}{{{x^4} - 1}} } = {\frac{{\frac{1}{4}}}{{x - 1}} - \frac{{\frac{1}{4}}}{{x + 1}} - \frac{{\frac{1}{2}}}{{{x^2} + 1}}.} \] So, the complete answer is \[ {\int {\frac{{dx}}{{{x^4} - 1}}} } = {\frac{1}{4}\int {\frac{{dx}}{{x - 1}}} - \frac{1}{4}\int {\frac{{dx}}{{x + 1}}} - \frac{1}{2}\int {\frac{{dx}}{{{x^2} + 1}}} } = {\frac{1}{4}\ln \left| {x - 1} \right| - \frac{1}{4}\ln \left| {x + 1} \right| - \frac{1}{2}\arctan x + C } = {\frac{1}{4}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2}\arctan x + C.} \]
   Example 9
Calculate the integral \({\large\int\normalsize} {{\large\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}}\normalsize} dx}.\)

Solution.
Decompose the integrand into partial functions, taking into account that the denominator has a third degree root: \[ {\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}} } = {\frac{A}{{{{\left( {x - 1} \right)}^3}}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{x - 1}}.} \] Equate coefficients: \[ {A + B\left( {x - 1} \right) + C{\left( {x - 1} \right)^2} = 5x,}\;\; {\Rightarrow A + Bx - B + C{x^2} - 2Cx + C = 5x,}\;\; {\Rightarrow C{x^2} + \left( {B - 2C} \right)x + A - B + C = 5x.} \] We get the following system of equations: \[ {\left\{ \begin{array}{l} C = 0\\ B - 2C = 5\\ A - B + C = 0 \end{array} \right.,}\;\; {\Rightarrow \left\{ \begin{array}{l} A = 5\\ B = 5\\ C = 0 \end{array} \right..} \] Hence, \[ {\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}} } = {\frac{5}{{{{\left( {x - 1} \right)}^3}}} + \frac{5}{{{{\left( {x - 1} \right)}^2}}}.} \] The initial integral is equal to \[ {\int {\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}}dx} } = {\int {\left( {\frac{5}{{{{\left( {x - 1} \right)}^3}}} + \frac{5}{{{{\left( {x - 1} \right)}^2}}}} \right)dx} } = {5\int {\frac{{dx}}{{{{\left( {x - 1} \right)}^3}}}} + 5\int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} } = {5 \cdot \frac{{{{\left( {x - 1} \right)}^{ - 2}}}}{{ - 2}} - \frac{5}{{x - 1}} + C } = { - \frac{5}{{2{{\left( {x - 1} \right)}^2}}} - \frac{5}{{x - 1}} + C.} \]
   Example 10
Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{{\left( {{x^2} + x - 1} \right)}^2}}}\normalsize} .\)

Solution.
Complete the square in the denominator \({{x^2} + x - 1}:\) \[ {\int {\frac{{dx}}{{{{\left( {{x^2} + x - 1} \right)}^2}}}} } = {\int {\frac{{dx}}{{{{\left( {{x^2} + x + \frac{1}{4} + \frac{3}{4}} \right)}^2}}}} } = {\int {\frac{{dx}}{{{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}^2}}}} .} \] Now, we can compute the integral using the reduction formula: \[ {\int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} } = {\frac{t}{{2{m^2}\left( {k - 1} \right){{\left( {{t^2} + {m^2}} \right)}^{k - 1}}}} } {+ \frac{{2k - 3}}{{2{m^2}\left( {k - 1} \right)}}\int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^{k - 1}}}}} } \] Then \[ {\int {\frac{{dx}}{{{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}^2}}}} } ={ \frac{1}{{2 \cdot \frac{3}{4} \cdot 1 \cdot \left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}} } +{ \frac{{4 - 3}}{{2 \cdot \frac{3}{4} \cdot 1}}\int {\frac{{dx}}{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}}} } ={ \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{2}{3}\int {\frac{{dx}}{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}}} } ={ \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{2}{3} \cdot \frac{2}{{\sqrt 3 }}\arctan \frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} + C } ={ \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{4}{{3\sqrt 3 }}\arctan \frac{{2x + 1}}{{\sqrt 3 }} + C.} \]

All Rights Reserved © www.math24.net, 2010-2016   info@math24.net