Question

$\int \frac{x^2+1}{x^4+1}dx$ will be equal to which of the following

• A. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c$
• B. $\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{2\sqrt{2}}\right)+c$
• C. $\frac{1}{5\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{3\sqrt{2}}\right)+c$
• D. $\frac{1}{5\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{5\sqrt{2}}\right)+c$

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c$

To solve these kind of integration problems we have to divide the numerator as well as the denominator by $x^2$ so that we can appropriately get the differentiation in numerator of what we substitute. Then we’ll make the suitable substitution to have a quadratic equation in the denominator. After that we can use the suitable standard formula.

So, let’s divide the numerator and denominator by $x^2$

$\int \frac{\frac{1}{x^2}+1}{x^2+\frac{1}{x^2}}dx$

$\text{Or}\int \frac{\frac{1}{x^2}+1}{{(x-\frac{1}{x})}^2+2}dx$

$\text{Now substitute}t=x-\frac{1}{x}$

$\text{So},dt=(1+\frac{1}{x^2})dx$

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

$=\int \frac{1}{(t{)}^2+2}dt=\int \frac{1}{(t{)}^2+(\sqrt{2}{)}^2}dt$

We can see that this is in standard form $\int \frac{1}{x^2+a^2}dx$ which is equal to

$\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)\text{So},\int \frac{1}{(t{)}^2+(\sqrt{2}{)}^2}dt=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Let’s substitute $t=x-\frac{1}{x}$ in the above equation. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)\text{Or}\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c$