Question

Evaluate: \[ \int_0^1 x^3\ln(1+x)\,dx \]

• A. \( \frac{7}{48} \)
• B. \( \frac{25}{48} \)
• C. \( \frac{1}{8} \)
• D. \( \frac{5}{24} \)

Hint:

For integrals containing logarithms: \[ \ln(\cdot) \] choose the logarithmic term as \(u\) during integration by parts. This usually simplifies the remaining integral significantly.

Answer 1

The Correct Option is A

Concept: Integrals involving logarithmic functions are commonly solved using integration by parts: \[ \int u\,dv=uv-\int v\,du \]

Step 1: Choose suitable functions for integration by parts.
Let: \[ u=\ln(1+x), \qquad dv=x^3dx \] Then: \[ du=\frac{1}{1+x}dx \] and \[ v=\frac{x^4}{4} \] Applying integration by parts: \[ I=\left[\frac{x^4}{4}\ln(1+x)\right]_0^1-\frac14\int_0^1 \frac{x^4}{1+x}dx \]

Step 2: Simplify the rational expression.
Using polynomial division: \[ \frac{x^4}{1+x}=x^3-x^2+x-1+\frac{1}{1+x} \] Thus: \[ I=\frac14\ln2-\frac14\int_0^1\left(x^3-x^2+x-1+\frac1{1+x}\right)dx \]

Step 3: Evaluate the integral term-by-term.
\[ \int_0^1 x^3dx=\frac14 \] \[ \int_0^1 x^2dx=\frac13 \] \[ \int_0^1 xdx=\frac12 \] \[ \int_0^1 1dx=1 \] \[ \int_0^1 \frac1{1+x}dx=\ln2 \] Substitute: \[ I=\frac14\ln2-\frac14\left(\frac14-\frac13+\frac12-1+\ln2\right) \] Simplify: \[ I=\frac14\ln2-\frac14\left(-\frac{7}{12}+\ln2\right) \] \[ I=\frac{7}{48} \] Therefore: \[ \boxed{\frac{7}{48}} \]