{x+1}{x≤ft(1+x e^{x})²} {dx}=

Correct Answer is : log⁡∣xex1+xex∣+11+xex+cl ≤ft|{x e^{x}}{1+x e^{x}}|+{1}{1+x e^{x}}+c_{l}log1+xexxex+1+xex1+cl (where c is a constant of integration)

≤ft|{x e^{x}}{1+x e^{x}}|+{x}{1+x e^{x}}+c_{l} (where c is a constant of integration)

≤ft|{x e^{x}}{1+x e^{x}}|+{e^{x}}{1+x e^{x}}+c_{l} (where c is a constant of integration)

≤ft|{x e^{x}}{1+x e^{x}}|+{1}{1+x e^{x}}+c_{l} (where c is a constant of integration)

≤ft|{x e^{x}}{1+x e^{x}}|-{1}{1+x e^{x}}+c_{l} (where c is a constant of integration)

{x+1}{x≤ft(1+x e^{x})²} {dx}=

Correct Answer is : log⁡∣xex1+xex∣+11+xex+cl ≤ft|{x e^{x}}{1+x e^{x}}|+{1}{1+x e^{x}}+c_{l}log1+xexxex+1+xex1+cl (where c is a constant of integration)

Test Index

MHTCET 11 May 2024 Shift 2 Paper

Question : 103 of 150

Marks: +1, -0

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

$\log \left|\frac{x e^{x}}{1+x e^{x}}\right|+\frac{x}{1+x e^{x}}+c_{l}$ (where c is a constant of integration)
$\quad \log \left|\frac{x e^{x}}{1+x e^{x}}\right|+\frac{e^{x}}{1+x e^{x}}+c_{l}$ (where c is a constant of integration)
$\log \left|\frac{x e^{x}}{1+x e^{x}}\right|+\frac{1}{1+x e^{x}}+c_{l}$ (where c is a constant of integration)
$\log \left|\frac{x e^{x}}{1+x e^{x}}\right|-\frac{1}{1+x e^{x}}+c_{l}$ (where c is a constant of integration)

Validate

Solution:

Concept:

Use the substitution

t = x e^x

and then apply partial fractions to integrate.

Explanation:

Let

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

Put

t = x e^x

Then

dt = (x+1)e^x dx

, so

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

Also

x(1+x e^x)^2 = t e^{-x}(1+t)^2

Therefore,

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

Now express

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

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

This gives

I = \log|t| - \log|1+t| + \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:

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

Hence, the correct option is C.

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