Calculus

Integration of Functions

Integration of Functions Logo

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 (that is 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 these steps in more details.

    Step 1. Reducing an Improper Fraction

    If the fraction is improper (that is 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

    \[
    {Q\left( x \right) }
    = {{\left( {x – a} \right)^\alpha } \cdots}\kern0pt{ {\left( {x – b} \right)^\beta }{\left( {{x^2} + px + q} \right)^\mu } \cdots}\kern0pt{ {\left( {{x^2} + rx + s} \right)^\nu },}
    \]

    where quadratic functions are irreducible, that is do not have real roots.

    Step 3. Decomposing the Rational Fraction into a Sum of Partial Fractions

    Write the function as follows:

    \[ {{\frac{{R\left( x \right)}}{{Q\left( x \right)}} }={ \frac{A}{{{{\left( {x – a} \right)}^\alpha }}} }\kern0pt{+ \frac{{{A_1}}}{{{{\left( {x – a} \right)}^{\alpha – 1}}}} + \ldots }}\kern0pt + {\frac{{{A_{\alpha – 1}}}}{{x – a}} + \ldots }\kern0pt + {\frac{B}{{{{\left( {x – b} \right)}^\beta }}} }+{ \frac{{{B_1}}}{{{{\left( {x – b} \right)}^{\beta – 1}}}} + \ldots }\kern0pt + {\frac{{{B_{\beta – 1}}}}{{x – b}} }\kern0pt + {\frac{{Kx + L}}{{{{\left( {{x^2} + px + q} \right)}^\mu }}} }+{ \frac{{{K_1}x + {L_1}}}{{{{\left( {{x^2} + px + q} \right)}^{\mu – 1}}}} + \ldots }\kern0pt + {\frac{{{K_{\mu – 1}}x + {L_{\mu – 1}}}}{{{x^2} + px + q}} + \ldots }\kern0pt + {\frac{{Mx + N}}{{{{\left( {{x^2} + rx + s} \right)}^\nu }}} }+{ \frac{{{M_1}x + {N_1}}}{{{{\left( {{x^2} + rx + s} \right)}^{\nu – 1}}}} + \ldots }\kern0pt + {\frac{{{M_{\nu – 1}}x + {N_{\nu – 1}}}}{{{x^2} + rx + s}}.} \]

    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:

    \[{1.\;\;}{\int {\frac{A}{{x – a}}dx} }={A \ln \left| {x – a} \right|}\]

    \[{2.\;\;}{\int {\frac{A}{{{{\left( {x – a} \right)}^k}}}dx} }={ \frac{A}{{\left( {1 – k} \right){{\left( {x – a} \right)}^{k – 1}}}} }\]

    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:

    \[{3.\;\;}{\int {\frac{{tdt}}{{{t^2} + {m^2}}}} }={ \frac{1}{2}\ln \left( {{t^2} + {m^2}} \right)}\]

    \[{4.\;\;}{ \int {\frac{{tdt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} }
    = {\frac{1}{{2\left( {1 – k} \right){{\left( {{t^2} + {m^2}} \right)}^{k – 1}}}} }
    \]

    \[{5.\;\;}{\int {\frac{{dt}}{{{t^2} + {m^2}}}} }={ \frac{1}{m}\arctan \frac{t}{m}}\]

    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:

    \[{6.\;\;}{ \int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} } = {\frac{t}{{2{m^2}\left( {k – 1} \right){{\left( {{t^2} + {m^2}} \right)}^{k – 1}}}} } + {\frac{{2k – 3}}{{2{m^2}\left( {k – 1} \right)}} }\kern0pt{ \int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^{k – 1}}}}} } \]


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Evaluate the integral \({\large\int\normalsize} {{\large\frac{{2x + 3}}{{{x^2} – 9}}\normalsize} dx}.\)

    Example 2

    Evaluate \({\large\int\normalsize} {{\large\frac{{{x^2} – 2}}{{x + 1}}\normalsize} dx}.\)

    Example 3

    Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{x^2} + 4x + 8}}\normalsize}.\)

    Example 4

    Evaluate the integral \({\large\int\normalsize} {\large\frac{{{x^2}dx}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)}}\normalsize}.\)

    Example 5

    Evaluate \({\large\int\normalsize} {\large\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}\normalsize}.\)

    Example 6

    Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{x^3} + 1}}\normalsize}.\)

    Example 7

    Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{x^4} – 1}}\normalsize}.\)

    Example 8

    Calculate the integral \({\large\int\normalsize} {{\large\frac{{5x}}{{{{\left( {x – 1} \right)}^3}}}\normalsize} dx}.\)

    Example 9

    Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{{{\left( {{x^2} + x – 1} \right)}^2}}}\normalsize}.\)

    Example 1.

    Evaluate the integral \({\large\int\normalsize} {{\large\frac{{2x + 3}}{{{x^2} – 9}}\normalsize} dx}.\)

    Solution.

    We 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

    \[
    {\int {\frac{{2x + 3}}{{{x^2} – 9}}dx} }
    = {{\frac{3}{2}\int {\frac{{dx}}{{x – 3}}} }+{ \frac{1}{2}\int {\frac{{dx}}{{x + 3}}} }}
    = {{\frac{3}{2}\ln \left| {x – 3} \right| }+{ \frac{1}{2}\ln \left| {x + 3} \right| }+{ C }}
    = {{\frac{1}{2}\ln \left| {{{\left( {x – 3} \right)}^3}\left( {x + 3} \right)} \right| }+{ C.}}
    \]

    Page 1
    Problem 1
    Page 2
    Problems 2-9