Question:
Evaluate: $\int \frac{\mathrm{x}^{2}+4 \mathrm{x}}{\mathrm{x}^{3}+6 \mathrm{x}^{2}+5} \mathrm{dx}$
Solution:
let $x^{3}+6 x^{2}+5=t$
Differentiating on both sides we get,
$\left(3 x^{2}+12 x\right) d x=d t$
$3\left(x^{2}+4 x\right) d x=d t$
$\left(x^{2}+4 x\right) d x=\frac{1}{3} d t$
Substituting it in $\int \frac{x^{2}+4 x}{x^{3}+6 x^{2}+5} d x$ we get,
$=\int \frac{1}{3 t} d t$
$=\frac{1}{3 \log \left(x^{3}+6 x^{2}+5\right)}+c$