question_answer

Evaluate \[\int{\frac{{{x}^{3}}}{{{x}^{4}}+3{{x}^{2}}+2}}\,dx.\]

Answer:

Let\[l=\int{\frac{{{x}^{3}}}{{{x}^{4}}+3{{x}^{2}}+2}}\,dx\] Put       \[{{x}^{2}}=t\] \[2xdx=dt\] \[xdx=\frac{1}{2}dt\] \[l=\frac{1}{2}\int{\frac{t}{{{t}^{2}}+3t+2}}\,dt\] \[=\frac{1}{2}\int{\frac{t}{{{t}^{2}}+2t+t+2}}\,dt\] \[=\frac{1}{2}\int{\frac{t}{(t+2)\,\,(t+1)}}\,dt\] Let\[\frac{t}{(t+1)\,\,(t+2)}=\frac{A}{(t+2)}+\frac{B}{(t+1)}\] \[=\frac{A\,(t+1)+B\,(t+2)}{(t+2)\,(t+1)}\] \[A+B=1\] and       \[A+2B=0\] So, \[A=2\]and \[B=-\,1\] Now, \[l=\frac{1}{2}\int{\frac{2}{t+2}}\,dt-\frac{1}{2}\int{\frac{1}{t+1}}\,dt\] \[=\log |t+2|-\frac{1}{2}\log |t+1|+C\] \[=\log |{{x}^{2}}+2|-\frac{1}{2}\log |{{x}^{2}}+1|+C\]