Question

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

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

Answer 1

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

Given, $\int f\left(x\right)dx=\psi \left(x\right)$
$I=\int x^{5}f\left(x^3\right)dx$
put $x^3=t\Rightarrow x^{2}dx=\frac{1}{3}dt$
$\Rightarrow I=\frac{1}{3}\int tf\left(t\right)dx$
$=\frac{1}{3}\left[t\int f\left(t\right)dt-\int \left(\int f\left(t\right)dt\right)dt\right]+c$
$=\frac{1}{3}\left[t\psi \right(t)-\int \left(\int f\left(t\right)dt\right)dt]+c$
$=\frac{1}{3}\left[x^3\psi \right(x^3)-\int 3x^2\psi \left(x^3\right)dx]+c$
$=\frac{1}{3}{x}^3\psi \left(x^3\right)-\int x^2\psi \left(x^3\right)dx+c$