Question

Evaluate the integral:\(\int\limits_0^1 {\frac{{{x^3} - 4x - 10}}{{{x^2} - x - 6}}} dx\)

Short answer

Hence, the value of the integral is \(\frac{{ - 1}}{5}\log \left( {\frac{{{3^{17}}}}{{{2^{18}}}}} \right)\)

Step-by-step solution

Step-1: Long division

Therefore,\(\frac{{{x^3} - 4x - 10}}{{{x^2} - x - 6}}\) Can be written as

\(\begin{aligned}{l}x + 1 + \frac{{( - 3x + 10)}}{{{x^2} - x - 6}}\\ = x + 1 + \frac{{( - 3x + 10)}}{{{x^2} - 3x + 2x - 6}}\\ = x + 1 + \frac{{( - 3x + 10)}}{{x(x - 3) + 2(x - 3)}}\\ = x + 1 + \frac{{( - 3x + 10)}}{{(x + 2)(x - 3)}}\end{aligned}\)

\(\int {(x + 1)dx + \int {\frac{{ - 3x + 10}}{{(x + 2)(x - 3)}}dx} } \)

Solving them separately

\(\begin{aligned}{l}\int {(x + 1)dx = \int {xdx + \int {1dx} } } \\ = \frac{{{x^2}}}{2} + x\end{aligned}\)

Step-2: Integrating the Equation

\(\frac{{( - 3x + 10)}}{{(x + 2)(x - 3)}}\)Can be written as

\(\begin{aligned}{l}\frac{A}{{x + 2}} + \frac{B}{{x - 3}}\\\frac{{( - 3x + 10)}}{{(x + 2)(x - 3)}} = \frac{{A(x - 3) + B(x + 2)}}{{(x + 2)(x - 3)}}\\ - 3x + 10 = A(x - 3) + B(x + 2)\\ - 3x + 10 = Ax - 3A - Bx + 2B\\ - 3x + 10 = (A + B)x - 3A + 2B\\A + B = - 3 \Rightarrow B = - 3 - A\\ - 3A + 2B = 10\\ - 3A + 2( - 3 - A) = 10\\ - 3A + ( - 6) - 2A = 10\\ - 5A = 16\\A = \frac{{ - 16}}{5}\\B = - 3 - A\\ = - 3 - \left( {\frac{{ - 16}}{5}} \right)\\ = \frac{{16}}{5} - 3 = \frac{{16 - 15}}{5} = \frac{1}{5}\end{aligned}\)

\(\int {\frac{{{\raise0.7ex\hbox{${ - 16}$} \!\mathord{\left/

{\vphantom {{ - 16} 5}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$5$}}}}{{x + 2}}} dx + \int {\frac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 5}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$5$}}}}{{x - 3}}dx} \)

Let\(x + 2\)be\(v\) and let\(x - 3\)be\(t\)

\(\begin{aligned}{l}x + 2 = v\\1 = \frac{{dv}}{{dx}}\\ \Rightarrow dx = dv\end{aligned}\) \(\begin{aligned}{l}x - 3 = t\\1 = \frac{{dt}}{{dx}}\\ \Rightarrow dx = dt\end{aligned}\)

\(\frac{{ - 16}}{5}\int {\frac{{dv}}{v}} \) and \(\frac{1}{5}\int {\frac{{dt}}{t}} \)

\(\begin{aligned}{l} = \frac{{ - 16}}{5}\log \left| v \right|\\ = \frac{{ - 16}}{5}\log \left| {x + 2} \right|\end{aligned}\) \(\begin{aligned}{l} = \frac{1}{5}\log \left| t \right|\\ = \frac{1}{5}\log \left| {x - 3} \right|\end{aligned}\)

\(\begin{aligned}{l}\frac{{ - 16}}{5}\log (x + 2) + \frac{1}{5}\log (x - 3)\\\frac{{ - 1}}{5}\left( {\log {{(x + 2)}^{16}} - \log (x + 3)} \right)\\\frac{{ - 1}}{5}\left( {\log \frac{{{{(x + 2)}^{16}}}}{{x + 3}}} \right)\end{aligned}\)

Step-3:  Putting the limits

\(\begin{aligned}{l}\frac{{ - 1}}{5}\left( {\log \frac{{{{(x + 2)}^{16}}}}{{x + 3}}} \right)\\\frac{{ - 1}}{5}\left( {\log \frac{{{{(1 + 2)}^{16}}}}{{(1 + 3)}} - \log \frac{{{2^{16}}}}{3}} \right)\\ = \frac{{ - 1}}{5}\left( {\log \frac{{{3^{16}}}}{4} - \log \frac{{{2^{16}}}}{3}} \right)\\ = \frac{{ - 1}}{5}\left( {\log \frac{{{3^{16}} \times 3}}{{4 \times {2^{16}}}}} \right)\\ = \frac{{ - 1}}{5}\log \left( {\frac{{{3^{17}}}}{{{2^{18}}}}} \right)\end{aligned}\)

Hence, the value of the integral is \(\frac{{ - 1}}{5}\log \left( {\frac{{{3^{17}}}}{{{2^{18}}}}} \right)\)