Q90 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
If $\int f(x)\, dx = g(x)$ then $\int x^3 f(x^2)\, dx$ is equal to
- Correct answerA$\dfrac{1}{2}\left[x^2 g(x^2) - \int g(x^2)\, d(x^2)\right]$
- B$\dfrac{1}{2}\left[x^2[f(x)]^2 - \int [g(x)]^2\, dx\right]$
- C$\dfrac{1}{2}\left[x^2 g(x) - \int g(x)\, d(x)\right]$
- D$\dfrac{1}{2}\left[x^2 g(x^2) + \int g(x^2)\, d(x^2)\right]$
Solution
Let $I = \int x^3 f(x^2)\, dx$.
Substitute $x^2 = t$, which gives $2x\, dx = dt$ or $x\, dx = \dfrac{1}{2} dt$.
$I = \int x^2 f(x^2) x\, dx = \dfrac{1}{2} \int t f(t)\, dt$
Using integration by parts, taking $t$ as the first function and $f(t)$ as the second function:
$I = \dfrac{1}{2} \left[ t \int f(t)\, dt - \int \left( \dfrac{d}{dt}(t) \int f(t)\, dt \right) dt \right]$
Since $\int f(x)\, dx = g(x)$, we have $\int f(t)\, dt = g(t)$.
$I = \dfrac{1}{2} \left[ t g(t) - \int g(t)\, dt \right]$
Substituting back $t = x^2$ and $dt = d(x^2)$:
$I = \dfrac{1}{2} \left[ x^2 g(x^2) - \int g(x^2)\, d(x^2) \right]$
Substitute $x^2 = t$, which gives $2x\, dx = dt$ or $x\, dx = \dfrac{1}{2} dt$.
$I = \int x^2 f(x^2) x\, dx = \dfrac{1}{2} \int t f(t)\, dt$
Using integration by parts, taking $t$ as the first function and $f(t)$ as the second function:
$I = \dfrac{1}{2} \left[ t \int f(t)\, dt - \int \left( \dfrac{d}{dt}(t) \int f(t)\, dt \right) dt \right]$
Since $\int f(x)\, dx = g(x)$, we have $\int f(t)\, dt = g(t)$.
$I = \dfrac{1}{2} \left[ t g(t) - \int g(t)\, dt \right]$
Substituting back $t = x^2$ and $dt = d(x^2)$:
$I = \dfrac{1}{2} \left[ x^2 g(x^2) - \int g(x^2)\, d(x^2) \right]$
Hence, the correct answer is option: 'A'.