Question

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

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

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+c$

Let’s divide the numerator and denominator by $x^2$

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

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

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

$dt=1+\frac{1}{x^2}$

$\int \frac{\frac{1}{x^2}+1}{{(x-\frac{1}{x})}^2+2}$ $Or\int \frac{1}{(t{)}^2+2}dtOr\int \frac{1}{(t{)}^2+(\sqrt{2}{)}^2}dt$

$\int \frac{1}{(x{)}^2+a^2}dx=\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)$
So, $\int \frac{1}{(t{)}^2+(\sqrt{2}{)}^2}dt=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{t}{\sqrt{2}}\right)+c$

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