{(x+1) d x}{x≤ft(1+x e^{x})}=

Correct Answer is : log⁡∣xex1+xex∣+c ≤ft|{x e^{x}}{1+x e^{x}}|+clog1+xexxex+c

≤ft|{(x+1) e^{x}}{x e^{x}}|+c

≤ft|{(x+1) e^{x}}{1+x e^{x}}|+c

Correct Answer is : log⁡∣xex1+xex∣+c ≤ft|{x e^{x}}{1+x e^{x}}|+clog1+xexxex+c

Test Index

MHTCET 18 Apr 2026 Shift 1 Paper

Question : 131 of 150

Marks: +1, -0

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

Solution:

Concept:

The integrand is the derivative of a logarithmic expression. We identify the function whose derivative gives the given integral.

Explanation:

Let

I = \int \frac{(x+1)\,dx}{x\left(1+xe^x\right)}

We know

\frac{d}{dx}\left(xe^x\right) = e^x(x+1)

Also,

\frac{d}{dx}\log\left|xe^x\right| = \frac{x+1}{x}

And

\frac{d}{dx}\log\left|1+xe^x\right| = \frac{e^x(x+1)}{1+xe^x}

Now consider

\log\left|\frac{xe^x}{1+xe^x}\right| = \log|xe^x| - \log|1+xe^x|

Differentiating, we get

\frac{x+1}{x} - \frac{e^x(x+1)}{1+xe^x}

Taking common denominator

x\left(1+xe^x\right)

\frac{(x+1)\left(1+xe^x\right) - xe^x(x+1)}{x\left(1+xe^x\right)} = \frac{x+1}{x\left(1+xe^x\right)}

This matches the given integrand exactly.

Therefore,

I = \log\left|\frac{xe^x}{1+xe^x}\right| + c

Answer:

Option A:

\log\left|\frac{xe^x}{1+xe^x}\right| + c

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