Question

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

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

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 $\int f\left(x\right)dx=\psi \left(x\right)\text{Let}I=\int x^{5}f\left(x^3\right)dx\text{Put}{x}^3=t\Rightarrow 3x^{2}dx=dtI=\frac{1}{3}\int 3.x^2.x^3.f\left(x^3\right).dx=\frac{1}{3}\int tf\left(t\right)dt=\frac{1}{3}\left[t\int f\left(t\right)dt-\int \left(\frac{d}{dt}\right(t)\int f(t\left)dt\right)dt\right]=\frac{1}{3}\left[t\psi \right(t)-\int \psi (t\left)dt\right]=\frac{1}{3}\left[x^3\psi \right(x^3)-3\int x^2\psi \left(x^3\right)dx+c]=\frac{1}{3}{x}^3\psi \left(x^3\right)-\int x^2\psi \left(x^3\right)dx+c$