Question

$\int \frac{x^{2}+1}{x^{4}+1}dx $

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

Answer 1

Answer: (D)

$ \int \frac{x^{2} +1}{x^{4} +1}dx = \int \frac{\left(1+ \frac{1}{x^{2}}\right)}{\left(x^{2} + \frac{1}{x^{2}}\right)} dx$
$ = \int \frac{\left(1 + \frac{1}{x^{2}}\right)}{\left(x- \frac{1}{x}\right)^{2} +2} dx $
Put $x - \frac{1}{x} =t \Rightarrow \left(1+ \frac{1}{x^{2}}\right) dx =dt $
$ = \int \frac{dt}{t^{2} +\left(\sqrt{2}\right)^{2}} = \frac{1}{\sqrt{2}}\tan^{-1} \left(\frac{t}{\sqrt{2}}\right) +c$
$ = \frac{1}{\sqrt{2}} \tan ^{-1} \left(\frac{x^{2} -1}{\sqrt{2} x}\right) +c $