Question

The value of \[ \int_0^1 x \log(1+x)\, dx \] (rounded off to two decimals) is __________.

Hint:

Whenever the integrand has $x\log(1+x)$, use integration by parts with $u=\log(1+x)$.

Answer 1

Correct Answer: 0.25

Use integration by parts: Let
\[ u = \log(1+x), \qquad dv = x\, dx. \]
Then
\[ du = \frac{1}{1+x} dx, \qquad v = \frac{x^2}{2}. \]
Apply integration by parts:
\[ \int_0^1 x\log(1+x)\, dx = \left[\frac{x^2}{2}\log(1+x)\right]_0^1 - \int_0^1 \frac{x^2}{2(1+x)}\, dx. \]
First term at $x=1$:
\[ \frac{1}{2}\log 2. \]
Zero at lower limit. Now simplify integrand:
\[ \frac{x^2}{2(1+x)} = \frac{1}{2}(x - 1 + \frac{1}{1+x}). \]
Thus:
\[ \int_0^1 x\log(1+x)\, dx = \frac{1}{2}\log 2 - \frac{1}{2}\left[\int_0^1 x\,dx - \int_0^1 dx + \int_0^1 \frac{1}{1+x}dx \right]. \]
Evaluate each:
\[ \int_0^1 x\,dx = \frac{1}{2}, \qquad \int_0^1 dx = 1, \qquad \int_0^1 \frac{1}{1+x}dx = \ln 2. \]
Substitute:
\[ I = \frac{1}{2}\ln 2 - \frac{1}{2}\left(\frac{1}{2} - 1 + \ln 2\right) = \frac{1}{2}\ln 2 - \frac{1}{2}\left(-\frac{1}{2} + \ln 2\right). \]
\[ I = \frac{1}{2}\ln 2 + \frac{1}{4} - \frac{1}{2}\ln 2 = \frac{1}{4}. \]
So:
\[ \boxed{0.25} \quad (\text{acceptable range: } 0.25\text{–}0.25) \]