Question

Let $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x,$ Then, for an arbitrary constant $C$, the value of $J-I$ equals

Accepted Answer

$\frac{1}{2} \log \left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)+C$

Answer

$\frac{1}{2} \log \left(\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right)+C$

Answer

$\frac{1}{2} \log \left(\frac{e^{2 x}+e^x+1}{e^{2 x}-e^x+1}\right)+C$

Answer

$\frac{1}{2} \log \left(\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right)+C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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