{1}{(e^{x}-1)²} dx =

Correct Answer is : log⁡[exex−1]−1ex−1+c ≤ft[ {e^{x}}{e^{x}-1} ] - {1}{e^{x}-1} + clog[ex−1ex]−ex−11+c

≤ft[ {e^{x}}{e^{x}-1} ] + {1}{e^{x}-1} + c

≤ft[ {e^{x}-1}{e^{x}} ] + {1}{e^{x}-1} + c

≤ft[ {e^{x}}{e^{x}-1} ] - {1}{e^{x}-1} + c

≤ft[ {e^{x}}{e^{x}-1} ] - {1}{e^{x}+1} + c

Correct Answer is : log⁡[exex−1]−1ex−1+c ≤ft[ {e^{x}}{e^{x}-1} ] - {1}{e^{x}-1} + clog[ex−1ex]−ex−11+c

Test Index

Integral Calculus

Question : 68 of 73

Marks: +1, -0

\int \frac{1}{(e^{x}-1)^{2}} dx =

Solution:

Concept:

Use substitution and partial fractions to integrate

1/(e^x-1)^2

Explanation:

Let

u = e^x

, so

du = e^x dx = u dx

, hence

dx = du/u

The integral becomes

\int \frac{1}{u(u-1)^2} \, du

Partial fractions:

\frac{1}{u(u-1)^2} = \frac{1}{u} - \frac{1}{u-1} + \frac{1}{(u-1)^2}

Integrate:

\ln|u| - \ln|u-1| - \frac{1}{u-1} + c = \ln\left|\frac{u}{u-1}\right| - \frac{1}{u-1} + c

Substitute back

u = e^x

\ln\left(\frac{e^x}{e^x-1}\right) - \frac{1}{e^x-1} + c

Answer:

\ln\left[\frac{e^x}{e^x-1}\right] - \frac{1}{e^x-1} + c

matches option C.

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