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 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: Reduce the fraction if it is improper (i.e. degree of $${P\left( x \right)}$$ is greater than degree of $${Q\left( x \right)}$$; Factor $${Q\left( x \right)}$$ into linear and/or quadratic (irreducible) factors; Decompose the fraction into a sum of partial fractions; 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: ${\frac{{R\left( x \right)}}{{Q\left( x \right)}} = \frac{A}{{{{\left( {x - a} \right)}^\alpha }}} + \frac{{{A_1}}}{{{{\left( {x - a} \right)}^{\alpha - 1}}}} + \ldots }\kern0pt {+ \frac{{{A_{\alpha - 1}}}}{{x - a}} + \ldots }\kern0pt {+ \frac{B}{{{{\left( {x - b} \right)}^\beta }}} + \frac{{{B_1}}}{{{{\left( {x - b} \right)}^{\beta - 1}}}} + \ldots }\kern0pt {+ \frac{{{B_{\beta - 1}}}}{{x - b}} }\kern0pt {+ \frac{{Kx + L}}{{{{\left( {{x^2} + px + q} \right)}^\mu }}} + \frac{{{K_1}x + {L_1}}}{{{{\left( {{x^2} + px + q} \right)}^{\mu - 1}}}} + \ldots }\kern0pt {+ \frac{{{K_{\mu - 1}}x + {L_{\mu - 1}}}}{{{x^2} + px + q}} + \ldots }\kern0pt {+ \frac{{Mx + N}}{{{{\left( {{x^2} + rx + s} \right)}^\nu }}} + \frac{{{M_1}x + {N_1}}}{{{{\left( {{x^2} + rx + s} \right)}^{\nu - 1}}}} + \ldots }\kern0pt {+ \frac{{{M_{\nu - 1}}x + {N_{\nu - 1}}}}{{{x^2} + rx + s}}.}$ 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 ${\int {\frac{{dx}}{{{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}^2}}}} } ={ \frac{1}{{2 \cdot \frac{3}{4} \cdot 1 \cdot \left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}} } +{ \frac{{4 - 3}}{{2 \cdot \frac{3}{4} \cdot 1}}\int {\frac{{dx}}{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}}} } ={ \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{2}{3}\int {\frac{{dx}}{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}}} } ={ \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{2}{3} \cdot \frac{2}{{\sqrt 3 }}\arctan \frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} + C } ={ \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{4}{{3\sqrt 3 }}\arctan \frac{{2x + 1}}{{\sqrt 3 }} + C.}$