Question

Evaluate the integral: \(\int\limits_1^2 {\frac{{4{y^2} - 7y - 12}}{{y(y + 2)(y - 3)}}dy} \)

Short answer

The solution of the given integralis as follows:

\(\frac{{g\ln (4) + 10\ln (2) - \ln (39366)}}{5} + c\)

Step-by-step solution

Step-1:  Split up the fraction & Integral

\(\begin{array}{l}\int\limits_1^2 {\frac{{4{y^2} - 7y - 12}}{{y(y + 2)(y - 3)}}dy} = \int\limits_1^2 {\frac{{4{y^2}}}{{y(y + 2)(y - 3)}}} dy - \int\limits_1^2 {\frac{{7y}}{{y(y + 2)(y - 3)}}dy - \int\limits_1^2 {\frac{{12}}{{y(y + 2)(y - 3)}}dy} } \\ = \int\limits_1^2 {\frac{{4y}}{{(y + 2)(y - 3)}}} dy - \int\limits_1^2 {\frac{7}{{(y + 2)(y - 3)}}dy} - \int\limits_1^2 {\frac{{12}}{{y(y + 2)(y - 3)}}dy} \end{array}\)

Step-2: Find the integral

\(\begin{array}{l}\int\limits_1^2 {\frac{{4{y^2} - 7y - 12}}{{y(y + 2)(y - 3)}}dy} = 4\int\limits_1^2 {\frac{y}{{(y + 2)(y + 3)}}dy - 7\int\limits_1^2 {\frac{1}{{(y + 2)(y + 3)}}} } dy - 12\int\limits_1^2 {\frac{1}{{y(y + 2)(y + 3)}}dy} \\ = 4\int\limits_1^2 {\frac{2}{{5(y + 2)}}} + \frac{3}{{5(y - 3)}}dy - 7\int\limits_1^2 {\frac{1}{{(y + 2)(y + 3)}}} dy + - 12\int\limits_1^2 {\frac{1}{{y(y + 2)(y + 3)}}} dy\\ = 4\left( {\frac{{2\ln \left( {\left| {y + 2} \right|} \right)}}{5} + \frac{{3\ln \left( {\left| {y - 3} \right|} \right)}}{5}dy} \right)_1^2 - 7\left( {\frac{{ - \ln \left( {\left| {y + 2} \right|} \right)}}{5} + \frac{{\ln \left( {\left| {y - 3} \right|} \right)}}{5}} \right)_1^2 - 12\left( { - \frac{{\ln \left( {\left| y \right|} \right)}}{6} + \frac{{\ln \left( {\left| {y + 2} \right|} \right)}}{{10}} + \frac{{\ln \left( {\left| {y - 3} \right|} \right)}}{{15}}} \right)_1^2\\ = \left( {\frac{{8\ln \left( {\left| {y + 2} \right|} \right)}}{5} + \frac{{12\ln \left( {\left| {y - 3} \right|} \right)}}{5} - 7\left( { - \frac{{\ln \left( {\left| {y + 2} \right|} \right)}}{5} + \frac{{\ln \left( {\left| {y - 3} \right|} \right)}}{5}} \right) + 2\ln \left( {\left| y \right|} \right) - \frac{{6\ln \left( {\left| {y + 2} \right|} \right)}}{5} - \frac{{4\ln \left( {\left| {y - 3} \right|} \right)}}{5}} \right)_1^2\end{array}\)

Step-3:  Simplify the expression by substituting the upper & lower limits

\(\begin{array}{l}\int\limits_1^2 {\frac{{4{y^2} - 7y - 12}}{{y(y + 2)(y - 3)}}dy} = \frac{{8\ln (12 + 21) + 12\ln (12 - 31)}}{5} - 7\left( { - \frac{{\ln (12 + 21)}}{5} + \frac{{\ln (12 - 31)}}{5}} \right) + 2\ln (121)\\ - \frac{{6\ln (12 + 21)}}{5} - \frac{{4\ln (12 - 31)}}{5}\\\\ = \left( {\frac{{8\ln (\left| {1 + 2} \right|) + 12\ln (1 - 3)}}{5} - 7\left( { - \frac{{\ln \left( {\left| {1 + 2} \right|} \right)}}{5} + \frac{{\ln \left( {\left| {1 - 3} \right|} \right)}}{5}} \right) + 2\ln \left( {\left| 1 \right|} \right) - \frac{{6\ln \left( {\left| {1 + 2} \right|} \right)}}{5} - 4\frac{{\ln \left( {\left| {1 - 3} \right|} \right)}}{5}} \right)\\ = \frac{{8\ln (4)}}{5} + \frac{{7\ln (4)}}{5} + 2\ln (2) - \frac{{6\ln (4)}}{5} + 0 - \frac{{\ln (39366)}}{5} + c\\ = \frac{{9\ln (4) + 10\ln (2) - \ln (39366)}}{5} + c\\\therefore \int\limits_1^2 {\frac{{4{y^2} - 7y - 12}}{{y(y + 2)(y - 3)}}dy} = \frac{{9\ln (4) + 10\ln (2) - \ln (39366)}}{5} + c\end{array}\)

Hence, the solution of the integral \(\int\limits_1^2 {\frac{{4{y^2} - 7y - 12}}{{y(y + 2)(y - 3)}}dy} \) is given as \(\frac{{9\ln (4) + 10\ln (2) - \ln (39366)}}{5} + c\)