What is the integral from-0.5 to 0 of xln(x+1) ?

The integral from-0.5 to 0 of xln(x+1) is 0.05256…

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integral from-0.5 to 0 of xln(x+1)

\int_{- 0.5}^{0} x ln (x + 1) d x

0.05256 \dots

Solution steps

\int_{- 0.5}^{0} x ln (x + 1) d x

Apply u-substitution

: \int_{0.5}^{1} (u - 1) ln (u) d u

= \int_{0.5}^{1} (u - 1) ln (u) d u

Apply Integration By Parts

: [ln (u) (0.5 u^{2} - u) - \int 0.5 u - 1 d u]_{0.5}^{1}

= [ln (u) (0.5 u^{2} - u) - \int 0.5 u - 1 d u]_{0.5}^{1}

\int 0.5 u - 1 d u = 0.25 u^{2} - u

= [ln (u) (0.5 u^{2} - u) - (0.25 u^{2} - u)]_{0.5}^{1}

Expand

ln (u) (0.5 u^{2} - u) - (0.25 u^{2} - u) : 0.5 u^{2} ln (u) - u ln (u) - 0.25 u^{2} + u

= [0.5 u^{2} ln (u) - u ln (u) - 0.25 u^{2} + u]_{0.5}^{1}

Compute the boundaries:

0.05256 \dots

= 0.05256 \dots

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What is the integral from-0.5 to 0 of xln(x+1) ?
The integral from-0.5 to 0 of xln(x+1) is 0.05256…