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. $\frac{1}{3}{x}^3\Psi \left(x^3\right)-\int x^2\Psi \left(x^3\right)dx+C$
• B. $\frac{1}{3}{x}^3\Psi \left(x^3\right)-3\int x^3\Psi \left(x^3\right)dx+C$
• C. $\frac{1}{4}{x}^3\Psi \left(x^3\right)-\int x^2\Psi \left(x^3\right)dx+C$
• D. $\frac{1}{3}[x^3\Psi \left(x^3\right)-\int x^3\Psi \left(x^3\right)dx]+C$

Answer 1

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

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

$=\int x^3\cdot x^{2}f\left(x^3\right)dx$

Put $x^3=t$, then $3x^{2}dx=dt$.

$I=\frac{1}{3}\int tf\left(t\right)dt$

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

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

$=\frac{1}{3}{x}^3\Psi \left(x^3\right)-\int x^2\Psi \left(x^3\right)dx$