Step 1 : Understanding the Question:
This integration problem asks us to evaluate a definite integral involving a product of an algebraic term (\( x^3 \)) and a transcendental logarithmic term (\( \ln(1+x) \)) over the interval \([0, 1]\).
Step 2 : Key Formulas and Approach:
The standard method to integrate products of different functional types is Integration by Parts:
\[
\int u \, dv = uv - \int v \, du
\]
We use the LIATE rule to choose \( u = \ln(1+x) \) and \( dv = x^3 dx \). After differentiation and integration of the respective parts, we will utilize polynomial division to simplify the resulting fraction.
Step 3 : Detailed Solution:
Choose the functions for integration by parts: let \( u = \ln(1+x) \) and \( dv = x^3 dx \).
Find the derivatives and integrals:
\[
du = \frac{1}{1+x}dx, \quad v = \frac{x^4}{4}
\]
Set up the integration by parts formula:
\[
I = \left[\frac{x^4}{4}\ln(1+x)\right]_0^1 - \int_0^1 \frac{x^4}{4(1+x)} dx
\]
Evaluate the boundary term at \( x = 1 \) and \( x = 0 \):
\[
\left(\frac{1^4}{4}\ln(2)\right) - \left(0\right) = \frac{1}{4}\ln 2
\]
Rewrite the integrand using algebraic division:
\[
\frac{x^4}{1+x} = x^3 - x^2 + x - 1 + \frac{1}{1+x}
\]
Integrate the expression term-by-term:
\[
\int_0^1 \left(x^3 - x^2 + x - 1 + \frac{1}{1+x}\right) dx = \left[ \frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} - x + \ln(1+x) \right]_0^1
\]
Substitute the limits:
\[
\left(\frac{1}{4} - \frac{1}{3} + \frac{1}{2} - 1 + \ln 2\right) = -\frac{7}{12} + \ln 2
\]
Combine the results to find the final integral \( I \):
\[
I = \frac{1}{4}\ln 2 - \frac{1}{4}\left(-\frac{7}{12} + \ln 2\right) = \frac{7}{48}
\]
Step 4 : Final Answer:
The value of the definite integral is \( \frac{7}{48} \), which corresponds to option (A).
\[
\boxed{\frac{7}{48}}
\]