Evaluate the following integrals:


Evaluate: $\int \frac{\mathrm{x}^{2}+4 \mathrm{x}}{\mathrm{x}^{3}+6 \mathrm{x}^{2}+5} \mathrm{dx}$


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,

$=\int \frac{1}{3 t} d t$

$=\frac{1}{3 \log \left(x^{3}+6 x^{2}+5\right)}+c$

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