Definite integral using substitution tech

Exercise:

Compute, $\int_{0}^{\frac{\pi }{4}}{\frac{\tan \theta {{\sec }^{2}}\theta }{{{\tan }^{2}}\theta +3\tan \theta +2}d\theta }$

Solution: Let $u=\tan \theta \Rightarrow du={{\sec }^{2}}\theta d\theta $ and $u\left( 0 \right)=0\,\,\And \,\,u\left( \frac{\pi }{4} \right)=1$

So \(\int_{0}^{\frac{\pi }{4}}{\frac{\tan \theta {{\sec }^{2}}\theta }{{{\tan }^{2}}\theta +3\tan \theta +2}d\theta }=\int_{0}^{1}{\frac{udu}{{{u}^{2}}+3u+2}}=\int_{0}^{1}{\frac{u}{\left( u+1 \right)\left( u+2 \right)}du}\)

Apply the partial fraction

$\frac{u}{\left( u+1 \right)\left( u+2 \right)}=\frac{A}{u+1}+\frac{B}{u+2}=\frac{Au+2A+Bu+B}{\left( u+1 \right)\left( u+2 \right)}=\frac{u\left( A+B \right)+2A+B}{\left( u+1 \right)\left( u+2 \right)}$

So $A+B=1\,\,\And \,\,2A+B=0$ $\Rightarrow A=-1\,\,\And \,\,B=2$

So $\int_{0}^{1}{\frac{u}{{{u}^{2}}+3u+2}du}=-\int_{0}^{1}{\frac{du}{u+1}+2\int_{0}^{1}{\frac{du}{u+2}}}$ $=\left[ -\ln \left| u+1 \right|+2\ln \left| u+2 \right| \right]_{0}^{1}$

                             $=-\ln 2+\ln 1+2\left( \ln 3-\ln 2 \right)=-3\ln 2+2\ln 3=\ln \left( \frac{9}{8} \right)$