Question:
If \(\displaystyle\int f(x)\,dx=\psi(x)\), then \(\displaystyle\int x^5 f(x^3)\,dx\) is equal to
If \(\displaystyle\int f(x)\,dx=\psi(x)\), then \(\displaystyle\int x^5 f(x^3)\,dx\) is equal to
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Write \(x^5dx=x^3\cdot x^2dx\), substitute \(t=x^3\), then integrate by parts.
Updated On: Nov 13, 2026
- \(\frac{1}{3}\left[x^3\psi(x^3)-\int x^2\psi(x^3)\,dx\right]+C\)
- \(\frac{1}{3}x^3\psi(x^3)-3\int x^3\psi(x^3)\,dx+C\)
- \(\frac{1}{3}x^3\psi(x^3)-\int x^2\psi(x^3)\,dx+C\)
- \(\frac{1}{3}\left[x^3\psi(x^3)-\int x^3\psi(x^3)\,dx\right]+C\)
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The Correct Option is C
Solution and Explanation
Let \(t=x^3\), so \(dt=3x^2dx\) and \(x^5dx=x^3\cdot x^2dx=\frac{t\,dt}{3}\).
\(\int x^5f(x^3)dx=\frac13\int t f(t)\,dt\).
By parts: \(\int tf(t)dt=t\psi(t)-\int\psi(t)dt\).
Back-substitute: \(\frac13\left[x^3\psi(x^3)-\int\psi(x^3)\,3x^2dx\right]=\frac13x^3\psi(x^3)-\int x^2\psi(x^3)dx+C\).
Answer: (3) \(\frac{1}{3}x^3\psi(x^3)-\int x^2\psi(x^3)\,dx+C\)
\(\int x^5f(x^3)dx=\frac13\int t f(t)\,dt\).
By parts: \(\int tf(t)dt=t\psi(t)-\int\psi(t)dt\).
Back-substitute: \(\frac13\left[x^3\psi(x^3)-\int\psi(x^3)\,3x^2dx\right]=\frac13x^3\psi(x^3)-\int x^2\psi(x^3)dx+C\).
Answer: (3) \(\frac{1}{3}x^3\psi(x^3)-\int x^2\psi(x^3)\,dx+C\)
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