{(x+1)(x+ x)²}{x} d x=

Correct Answer is : (x+log⁡x)33+c{(x+ x)³}{3}+c3(x+logx)3+c, where c is the constant of integration

≤ft(x+ x/x)²+c, where c is the constant of integration

{(x+ x)²}{x}+c, where c is the constant of integration

{(x+ x)³}{3}+c, where c is the constant of integration

{(x+ x)³}{3 x}+c, where c is the constant of integration

Correct Answer is : (x+log⁡x)33+c{(x+ x)³}{3}+c3(x+logx)3+c, where c is the constant of integration

Test Index

MHTCET 16 Apr 2026 Shift 1 Paper

Question : 126 of 150

Marks: +1, -0

\int \frac{(x+1)(x+\log x)^{2}}{x} d x=

$\left(\frac{x+\log x}{x}\right)^{2}+c$ , where c is the constant of integration
$\frac{(x+\log x)^{2}}{x}+c$ , where c is the constant of integration
$\frac{(x+\log x)^{3}}{3}+c$ , where c is the constant of integration
$\frac{(x+\log x)^{3}}{3 x}+c$ , where c is the constant of integration

Solution:

Concept:

Use substitution by recognising the derivative of

x+\log x

in the integrand.

Explanation:

The given integral is:

\int \frac{(x+1)(x+\log x)^{2}}{x}\,dx

Rewrite the fraction:

\frac{x+1}{x} = 1+\frac{1}{x}

Therefore, the integral becomes:

\int (x+\log x)^{2}\left(1+\frac{1}{x}\right)dx

Now, differentiate

x+\log x

\frac{d}{dx}(x+\log x) = 1+\frac{1}{x}

Substitute

u = x+\log x

, so

du = \left(1+\frac{1}{x}\right)dx

The integral simplifies to:

\int u^{2}\,du = \frac{u^{3}}{3}+c

Substitute

u = x+\log x

back:

\frac{(x+\log x)^{3}}{3}+c

This matches Option C.

Answer:

\frac{(x+\log x)^{3}}{3}+c

Hence, Option C is correct.

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