Question

Evaluate: $\int \frac{dx}{(x+1)\sqrt{2x^2+3x+1}}$.

Answer 1

$\int \frac{dx}{(x+1)\sqrt{2x^2+3x+1}}$

$=\int \frac{dx}{{(x+1)}^{\frac{3}{2}}\sqrt{(2x+1)}}$

Let

$u=\sqrt{2x+1}\Rightarrow x=\frac{u^2-1}{2}$

Differentiating above equation w.r.t. $x$, we have

$du=\frac{dx}{\sqrt{2x+1}}$

Thus the integral will be in form-

$\int \frac{du}{{\left(\frac{u^2+1}{2}\right)}^{\frac{3}{2}}}$

$=\int \frac{2\sqrt{2}du}{{(u^2+1)}^{\frac{3}{2}}}$

$=2\sqrt{2}\int \frac{du}{{(u^2+1)}^{\frac{3}{2}}}$

Now,

Let $u=\tan t\Rightarrow du={\sec }^{2}tdt$

Now the integral will become,

$2\sqrt{2}\int \frac{{\sec }^{2}dt}{{({\tan }^{2}t+1)}^{\frac{3}{2}}}$

$=2\sqrt{2}\int \frac{{\sec }^{2}tdt}{{\left({\sec }^{2}t\right)}^{\frac{3}{2}}}$

$=2\sqrt{2}\int \frac{{\sec }^{2}tdt}{{\sec }^{3}t}$

$=2\sqrt{2}\int \cos tdt$

$=2\sqrt{2}\sin t+C$

$=2\sqrt{2}\sin \left({\tan }^{-1}u\right)+C$

$=2\sqrt{2}\frac{u}{\sqrt{1+u^2}}$

$=2\sqrt{2}\frac{\sqrt{2x+1}}{\sqrt{1+{\left(\sqrt{2x+1}\right)}^2}}$

$=\frac{2\sqrt{2x+1}}{\sqrt{x+1}}$

$=2\sqrt{\frac{2x+1}{x+1}}$

Hence

$\int \frac{dx}{(x+1)\sqrt{2x^2+3x+1}}=2\sqrt{\frac{2x+1}{x+1}}$