Statement-1: The value of the integral $\int \frac{e^{3 x}+e^x}{e^{4 x}+1} d x$ is $\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{e^x-e^{-x}}{\sqrt{2}}\right)+C$
Statement-2: A primitive of the function $f(x)=\frac{x^2+1}{x^4+1}$ is $\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)$
Answer & explanation
Correct answer: option 1
A primitive of the function $f(x)=\frac{x^2+1}{x^4+1}$ is given by
$I=\int \frac{x^2+1}{x^4+1} d x=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}} d x$
$\Rightarrow I=\int \frac{1}{\left(x-\frac{1}{x}\right)^2+(\sqrt{2})^2} d\left(x-\frac{1}{x}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C$
$\Rightarrow I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)+C$
So, statement-2 is true.
Now, $I=\int \frac{e^{3 x}+e^x}{e^{4 x}+1} d x=\int \frac{\left(e^x\right)^2+1}{\left(e^x\right)^4+1} d\left(e^x\right)$
$\Rightarrow I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{e^{2 x}-1}{\sqrt{2} e^x}\right)+C$ [Using statement-2]
$\Rightarrow I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{e^x-e^{-x}}{\sqrt{2}}\right)+C$
So, statement- 1 is true. Also, statement-2 is a correct explanation for statement-1.