If I=\int {e^x}/{e^{4 x}+e^{2 x}+1} \dx and J=\int {e^{-x}}/{e^{-4 x}+e^{-2 x}+1} \dx, then for any arbitrary constant C, than the value of J-I equals

Correct Answer is : 12log|(e2x−ex+1e2x+ex+1)|+C

\frac{1}{2} \log {| ( {e^{4 x}-e^{2 x}+1}/{e^{4 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^{2 x}-e^x+1}/{e^{2 x}+e^x+1}) |}+C

\frac{1}{2} \log {| ( {e^{4 x}+e^{2 x}+1}/{e^{4 x}-e^{2 x}+1}) |}+C

MHTCET 13 May 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 93 of 150

Marks: +1, -0

If I=∫‌

e4x+e2x+1

‌dx and J=∫‌

e−x

e−4x+e−2x+1

‌dx, then for any arbitrary constant C, than the value of J−I equals

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