Statement-1: If $I_1=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x$ and
$I_2=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$, then $I_2-I_1=\frac{1}{2} \log \left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)+C$ where C is an arbitrary constant.
Statement-2: A primitive of $f(x)=\frac{x^2-1}{x^4+x^2+1}$ is $\frac{1}{2} \log \left(\frac{x^2-x+1}{x^2+x+1}\right)$
Answer & explanation
Correct answer: option 1
A primitive of $f(x)=\frac{x^2-1}{x^4+x^2+1}$ is given by
$I =\int \frac{x^2-1}{x^4+x^2+1} d x=\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} d x$
$\Rightarrow I =\int \frac{1}{\left(x+\frac{1}{x}\right)^2-1^2} d\left(x+\frac{1}{x}\right)=\frac{1}{2} \log \mid \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1 \mid}+C$
$\Rightarrow I =\frac{1}{2} \log \left(\frac{x^2-x+1}{x^2+x+1}\right)+C$
So, statement- 2 is true.
Now, $I_2-I_1=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x-\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x$
$\Rightarrow I_2-I_1=\int \frac{e^{3 x}}{e^{4 x}+e^{2 x}+1} d x-\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x$
$\Rightarrow I_2-I_1=\int \frac{e^{2 x}-1}{e^{4 x}+e^{2 x}+1} d\left(e^x\right)$
$\Rightarrow I_2-I_1=\frac{1}{2} \log \left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)+C$ [Using statement-2]
So, statement- 1 is true. Also, statement-2 is a correct explanation for statement-1.