Question:
\(\int x^3logx\,dx =\)
\(\int x^3logx\,dx =\)
Show Hint
Use integration by parts with \(u=\log x\) and \(dv=x^3dx\).
Updated On: Oct 1, 2026
- \(\frac{x^4}{16}[4logx-1]+c\)
- \(\frac{x^4}{16}[4logx+1]+c\)
- \(\frac{x^4}{16}[-4logx-1]+c\)
- \(\frac{x^4}{16}[-4logx+1]+c\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Integration by parts: \(\int u\,dv=uv-\int v\,du\).
Step 2: Key Formula or Approach
Take \(u=\log x\), \(dv=x^3dx\), so \(du=dx/x\) and \(v=x^4/4\).
Step 3: Detailed Explanation
\[ \int x^3\log x\,dx=\frac{x^4}{4}\log x-\int\frac{x^3}{4}\,dx=\frac{x^4}{4}\log x-\frac{x^4}{16}+c \]
\[ =\frac{x^4}{16}\left[4\log x-1\right]+c \]
Final Answer:
The integral is \(\frac{x^4}{16}[4\log x-1]+c\), option (A). \[ \boxed{\dfrac{x^4}{16}\left[4\log x-1\right]+c\ \text{(A)}} \]
Integration by parts: \(\int u\,dv=uv-\int v\,du\).
Step 2: Key Formula or Approach
Take \(u=\log x\), \(dv=x^3dx\), so \(du=dx/x\) and \(v=x^4/4\).
Step 3: Detailed Explanation
\[ \int x^3\log x\,dx=\frac{x^4}{4}\log x-\int\frac{x^3}{4}\,dx=\frac{x^4}{4}\log x-\frac{x^4}{16}+c \]
\[ =\frac{x^4}{16}\left[4\log x-1\right]+c \]
Final Answer:
The integral is \(\frac{x^4}{16}[4\log x-1]+c\), option (A). \[ \boxed{\dfrac{x^4}{16}\left[4\log x-1\right]+c\ \text{(A)}} \]
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