Question

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

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

Answer 1

The correct option is C

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

Explanation for the correct option.

Find the value of the given integral.

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

Now using the substitution the integral $\int x^{5}f\left(x^3\right)dx$ can be simplified as:

$\begin{array}{rcl}\int x^{5}f\left(x^3\right)dx& =& \int x^3{x}^{2}f\left(x^3\right)dx\\ & =& \int \frac{1}{3}{x}^{3}f\left(x^3\right)\left(3x^{2}dx\right)\\ & =& \frac{1}{3}\int t·f\left(t\right)dt\mathbf{}\mathbf{[}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{t}\mathbf{,}\mathbf{}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{d}\mathbf{t}\mathbf{]}\\ & =& \frac{1}{3}\left[t\int f\left(t\right)dt-\int \left(1\times \int f\left(t\right)dt\right)dt\right]+C\\ & =& \frac{1}{3}\left[t\psi \left(t\right)-\int \psi \left(t\right)dt\right]+C\mathbf{[}\mathbf{\int }\mathbf{f}\left(x\right)\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{\psi }\left(x\right)\mathbf{]}\\ & =& \frac{1}{3}\left[x^3\psi \left(x^3\right)-\int \psi \left(x^3\right)3x^{2}dx\right]+C\mathbf{[}{\mathit{x}}^{\mathbf{3}}\mathbf{=}\mathit{t}\mathbf{,}\mathbf{}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathit{d}\mathit{x}\mathbf{=}\mathit{d}\mathit{t}\mathbf{]}\\ & =& \frac{1}{3}{x}^3\psi \left(x^3\right)-\int x^2\psi \left(x^3\right)dx+C\end{array}$

Thus the value of the integral $\int x^{5}f\left(x^3\right)dx$ is $\frac{1}{3}{x}^3\psi \left(x^3\right)-\int x^2\psi \left(x^3\right)dx+C$.

Hence, the correct option is C.