int_0^1 (1+x) log (1+x) , dx = | vedclass

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(D) Let $I = \int_0^1 (1+x) \log (1+x) \, dx$. Using integration by parts,let $u = \log(1+x)$ and $dv = (1+x) \, dx$. Then $du = \frac{1}{1+x} \, dx$

Vedclass · Accepted Answer

$\frac{-3}{4} + 2 \log 2$

Vedclass · Answer

(D) Let $I = \int_0^1 (1+x) \log (1+x) \, dx$. Using integration by parts,let $u = \log(1+x)$ and $dv = (1+x) \, dx$. Then $du = \frac{1}{1+x} \, dx$ and $v = \frac{(1+x)^2}{2}$. $I = \left[ \frac{(1+x)^2}{2} \log(1+x) \right]_0^1 - \int_0^1 \frac{(1+x)^2}{2} \cdot \frac{1}{1+x} \, dx$ $I = \left[ \frac{4}{2} \log 2 - 0 \right] - \frac{1}{2} \int_0^1 (1+x) \, dx$ $I = 2 \log 2 - \frac{1}{2} \left[ x + \frac{x^2}{2} \right]_0^1$ $I = 2 \log 2 - \frac{1}{2} \left( 1 + \frac{1}{2} \right) = 2 \log 2 - \frac{1}{2} \left( \frac{3}{2} \right)$ $I = 2 \log 2 - \frac{3}{4} = \frac{-3}{4} + 2 \log 2$.

$\int_0^1 (1+x) \log (1+x) \, dx =$

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