Question

If $\int f\left(x\right)dx$ $=\Psi \left(x\right)$ , then $\int x^{5}f\left(x^3\right)dx$ is equal to

• A. $\frac{1}{3}{x}^3\Psi \left(x^3\right)-3\int x^3\Psi \left(x^3\right)dx+C$
• B. $\frac{1}{3}{x}^3\Psi \left(x^3\right)-\int x^2\Psi \left(x^3\right)dx+C$
• C. $\frac{1}{3}\left[x^3\Psi \right(x^3)-\int x^3\Psi (x^3\left)dx\right]+C$
• D. $\frac{1}{3}\left[x^3\Psi \right(x^3)-\int x^2\Psi (x^3\left)dx\right]+C$

Answer 1

The correct option is B $\frac{1}{3}{x}^3\Psi \left(x^3\right)-\int x^2\Psi \left(x^3\right)dx+C$

$\int f\left(x\right)dx=\Psi \left(x\right)$

Let $x^3=t$

$\Rightarrow 3x^{2}dx=dt$

Then

$\int x^{5}f\left(x^3\right)dx$

$=\frac{1}{3}\int tf\left(t\right)dt$....................(replacing the value of $x^3$ & $dx$ )

$=\frac{1}{3}[t\int f(t)dt-\int \{1\cdot \int f\left(t\right)dt\left\}dt\right]$

$=\frac{1}{3}\left[t\Psi \right(t)-\int \Psi (t)dt+C].$

$=\frac{1}{3}\left[x^3\Psi \right(x^3)-\int x^2\Psi (x^3\left)dx\right]+C.$