{(x+1)}{x≤ft(1+x e^{x})²} d x=

Correct Answer is : log⁡(xex1+xex)+1(1+xex)+c ≤ft({x e^{x}}{1+x e^{x}})+{1}{≤ft(1+x e^{x})}+clog(1+xexxex)+(1+xex)1+c

_{- }≤ft({x e^{x}}{1+x e^{x}})+{1}{≤ft(1+x e^{x})}+c

≤ft({x e^{x}}{1+x e^{x}})+{1}{≤ft(1+x e^{x})}+c

≤ft({1+x e^{x}}{x e^{x}})+≤ft(1+x e^{x})+c

- ≤ft({x e^{x}}{1+x e^{x}})-{1}{≤ft(1+x e^{x})}+c

Correct Answer is : log⁡(xex1+xex)+1(1+xex)+c ≤ft({x e^{x}}{1+x e^{x}})+{1}{≤ft(1+x e^{x})}+clog(1+xexxex)+(1+xex)1+c

Test Index

MHTCET 13 Apr 2026 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x=

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

Solution:

Concept:

Use substitution

t = x e^{x}

and then apply partial fractions.

Explanation:

Let

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

Put

t = x e^{x}

Then

\frac{dt}{dx} = e^{x}(x+1)

dt = e^{x}(x+1) dx

Divide both sides by

t = x e^{x}

\frac{dt}{t} = \frac{x+1}{x} dx

The integral becomes:

I = \int \frac{1}{(1+t)^{2}} \cdot \frac{dt}{t} = \int \frac{dt}{t(1+t)^{2}}

Using partial fractions:

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

Integrate term by term:

I = \int \left(\frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^{2}}\right) dt

I = \log t - \log(1+t) + \frac{1}{1+t} + c

I = \log\left(\frac{t}{1+t}\right) + \frac{1}{1+t} + c

Substitute back

t = x e^{x}

I = \log\left(\frac{x e^{x}}{1+x e^{x}}\right) + \frac{1}{1+x e^{x}} + c

Answer:

Option B:

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

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