Question

Assertion :$\int \left(\frac{1}{1+x^4}\right)dx={\tan }^{-1}\left(x^2\right)+C$ Reason: $\int \frac{1}{1+x^2}dx={\tan }^{-1}x+C$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect and Reason is correct

Answer 1

The correct option is D Assertion is incorrect and Reason is correct

Assertion Statement:$I=\int \frac{1}{x^4+1}dx$
$=\int \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=\frac{1}{2}\int \frac{\frac{2}{x^2}}{x^2+\frac{1}{x^2}}dx$
$=\frac{1}{2}\int (\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}-\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}})dx$
$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$
$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+2}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-2}$
Substituting, $x-\frac{1}{x}=u$ in first integral $\Rightarrow (1+\frac{1}{x^2})dx=du$
And $x+\frac{1}{x}=v$ in second integral $\Rightarrow (1-\frac{1}{x^2})dx=dv$, we get
$I=\frac{1}{2}\int \frac{du}{u^2+{\left(\sqrt{2}\right)}^2}-\frac{1}{2}\int \frac{dv}{v^2-{\left(\sqrt{2}\right)}^2}$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)-\frac{1}{2}\times \frac{1}{2\sqrt{2}}\log \left|\frac{v-\sqrt{2}}{\nu +\sqrt{2}}\right|+C$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{4\sqrt{2}}\log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)-\frac{1}{4\sqrt{2}}\log \left|\frac{x^2-\sqrt{2}x+1}{x^2+x\sqrt{2}+1}\right|+C$

Hence, Assertion is incorrect.
Reason statement:
$\int \frac{1}{1+x^2}dx={\tan }^{-1}x+C$
Consider, $\int \frac{1}{1+x^2}dx$
Put $x=\tan \theta \Rightarrow dx={\sec }^2\theta$
$\int \frac{{\sec }^2\theta }{1+{\tan }^2\theta }d\theta =\int d\theta$
$=\theta +C={\tan }^{-1}x+C$
Hence, the reason is correct.