If {\I}_1=\int {e^x}/{e^{4 x}+e^{2 x}+1} \dx, {\I}_2=\int {e^{-x}}/{e^{-4 x}+e^{-2 x}+1} \dx, then {\I}_2- {\I}_1=

Correct Answer is : 12log(ex+e−x−1ex+e−x+1)+c

\frac{1}{2} \log ( {e^{2 x}-e^{-2 x}+1}/{e^{2 x}+e^{-2 x}-1}) + {\c}

\frac{1}{2} \log ( {e^{2 x}-e^{-2 x}-1}/{e^{2 x}+e^{-2 x}+1}) + {\c}

\frac{1}{2} \log ( {e^{2 x}+e^{-x}+1}/{e^{2 x}+e^{-x}-1}) + {\c}

\frac{1}{2} \log ( {e^x+e^{-x}-1}/{e^x+e^{-x}+1}) + {\c}

TG EAMCET 4 May 2025 Shift 2 Paper

Question : 72 of 160

Marks: +1, -0

If I1=∫‌

e4x+e2x+1

‌dx,I2=∫‌

e−x

e−4x+e−2x+1

‌dx, then I2−I1=

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