What is the integral from 0 to 1/4 of xln(x+1) ?

The integral from 0 to 1/4 of xln(x+1) is (-30ln(5/4)+55)/(64)-3/4

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integral from 0 to 1/4 of xln(x+1)

\int_{0}^{\frac{1}{4}} x ln (x + 1) d x

\frac{- 30 ln ( \frac{5}{4} ) + 55}{64} - \frac{3}{4}

Solution steps

\int_{0}^{\frac{1}{4}} x ln (x + 1) d x

Apply u-substitution

: \int_{1}^{\frac{5}{4}} (u - 1) ln (u) d u

= \int_{1}^{\frac{5}{4}} (u - 1) ln (u) d u

Apply Integration By Parts

: [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{1}^{\frac{5}{4}}

= [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{1}^{\frac{5}{4}}

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{\frac{5}{4}}

Simplify

[ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{\frac{5}{4}} : [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{1}^{\frac{5}{4}}

= [\frac{1}{4} (2 u^{2} ln (u) - 4 u ln (u) - u^{2} + 4 u)]_{1}^{\frac{5}{4}}

Compute the boundaries:

\frac{- 30 ln ( \frac{5}{4} ) + 55}{64} - \frac{3}{4}

= \frac{- 30 ln ( \frac{5}{4} ) + 55}{64} - \frac{3}{4}

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What is the integral from 0 to 1/4 of xln(x+1) ?
The integral from 0 to 1/4 of xln(x+1) is (-30ln(5/4)+55)/(64)-3/4