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Question

Mathematics Question on integral

∫x2+1x4+1dx\int \frac{x^{2}+1}{x^{4}+1}dx

A

12log⁡e(x2+1)+c\frac{1}{\sqrt{2} } \log_{e}\left(x^{2} +1\right)+c

B

12tan⁡−1(x2+1x2)+c\frac{1}{\sqrt{2} } \tan^{-1} \left(\frac{x^{2}+1}{x\sqrt{2}}\right)+c

C

12tan⁡−1(x2+1)+c\frac{1}{\sqrt{2} } \tan^{-1} \left(x^{2} +1\right)+c

D

12tan⁡−1(x2−1x2)+c\frac{1}{\sqrt{2} } \tan^{-1} \left(\frac{x^{2} - 1}{x\sqrt{2}}\right)+c

Answer

12tan⁡−1(x2−1x2)+c\frac{1}{\sqrt{2} } \tan^{-1} \left(\frac{x^{2} - 1}{x\sqrt{2}}\right)+c

Explanation

Solution

∫x2+1x4+1dx=∫(1+1x2)(x2+1x2)dx \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
=∫(1+1x2)(x−1x)2+2dx= \int \frac{\left(1 + \frac{1}{x^{2}}\right)}{\left(x- \frac{1}{x}\right)^{2} +2} dx
Put x−1x=t⇒(1+1x2)dx=dtx - \frac{1}{x} =t \Rightarrow \left(1+ \frac{1}{x^{2}}\right) dx =dt
=∫dtt2+(2)2=12tan⁡−1(t2)+c= \int \frac{dt}{t^{2} +\left(\sqrt{2}\right)^{2}} = \frac{1}{\sqrt{2}}\tan^{-1} \left(\frac{t}{\sqrt{2}}\right) +c
=12tan⁡−1(x2−12x)+c= \frac{1}{\sqrt{2}} \tan ^{-1} \left(\frac{x^{2} -1}{\sqrt{2} x}\right) +c