Q91 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
$\displaystyle\int \dfrac{(x + 1)(x + \log x)^2}{x}\,dx =$
- A$\left(\dfrac{x + \log x}{x}\right)^2 + c$, where c is the constant of integration
- B$\dfrac{(x + \log x)^2}{x} + c$, where c is the constant of integration
- Correct answerC$\dfrac{(x + \log x)^3}{3} + c$, where c is the constant of integration
- D$\dfrac{(x + \log x)^3}{3x} + c$, where c is the constant of integration
Solution
Let $I = \displaystyle\int \dfrac{(x + 1)(x + \log x)^2}{x} \, dx$
$I = \displaystyle\int \left( 1 + \dfrac{1}{x} \right) (x + \log x)^2 \, dx$
Substituting $x + \log x = t$
Differentiating with respect to $x$,
$\left( 1 + \dfrac{1}{x} \right) dx = dt$
Substituting $t$ and $dt$ into the integral,
$I = \displaystyle\int t^2 \, dt$
$I = \dfrac{t^3}{3} + c$
Substituting back $t = x + \log x$,
$I = \dfrac{(x + \log x)^3}{3} + c$
Answer: $\dfrac{(x + \log x)^3}{3} + c$, where c is the constant of integration
$I = \displaystyle\int \left( 1 + \dfrac{1}{x} \right) (x + \log x)^2 \, dx$
Substituting $x + \log x = t$
Differentiating with respect to $x$,
$\left( 1 + \dfrac{1}{x} \right) dx = dt$
Substituting $t$ and $dt$ into the integral,
$I = \displaystyle\int t^2 \, dt$
$I = \dfrac{t^3}{3} + c$
Substituting back $t = x + \log x$,
$I = \dfrac{(x + \log x)^3}{3} + c$
Answer: $\dfrac{(x + \log x)^3}{3} + c$, where c is the constant of integration
Hence, the correct answer is option: 'C'.