Question

$\int \frac{(x^2-1)dx}{(x^2+1)\sqrt{x^4+1}}$ is equal to :

• A. ${\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2x}}\right)+c$
• B. $\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2x}}\right)+c$
• C. $\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2}}\right)+c$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

We have,

$\int \frac{(x^2-1)dx}{(x^2+1)\sqrt{x^4+1}}$

Now,

Let

$t=\frac{x}{\sqrt{x^4+1}}$

$dt=\frac{1-x^4}{{(x^4+1)}^{\frac{3}{2}}}dx$

So,

$\int \frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx=\int \frac{(x^2-1)(x^2+1)(x^4+1)}{{(x^2+1)}^2{(x^4+1)}^{\frac{3}{2}}}dx$

$=\int \frac{1}{1+2t^2}.\frac{x^4-1}{{(x^4+1)}^{\frac{3}{2}}}dx$

$=-\int \frac{dt}{1+2t^2}$

On integration and we get,

$-\int \frac{dt}{1+2t^2}=-\frac{1}{\sqrt{2}}{\tan }^{-1}\sqrt{2}t$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x\sqrt{2}}{\sqrt{x^4+1}}+C$

Hence, this is the answer.