What is the integral from a to 1 of x^pln(x) ?

The integral from a to 1 of x^pln(x) is (-1+a^{p+1})/((1+p)^2)-(a^{p+1}ln(a))/(p+1)

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integral from a to 1 of x^pln(x)

\int_{a}^{1} x^{p} ln (x) d x

\frac{- 1 + a ^{p + 1}}{( 1 + p ) ^{2}} - \frac{a ^{p + 1} ln ( a )}{p + 1}

Solution steps

\int_{a}^{1} x^{p} ln (x) d x

Apply Integration By Parts

: [\frac{1}{p + 1} x^{p + 1} ln (x) - \int \frac{x ^{p}}{p + 1} d x]_{a}^{1}

= [\frac{1}{p + 1} x^{p + 1} ln (x) - \int \frac{x ^{p}}{p + 1} d x]_{a}^{1}

\int \frac{x ^{p}}{p + 1} d x = \frac{x ^{p + 1}}{( p + 1 ) ^{2}}

= [\frac{1}{p + 1} x^{p + 1} ln (x) - \frac{x ^{p + 1}}{( p + 1 ) ^{2}}]_{a}^{1}

Compute the boundaries:

- \frac{1}{( p + 1 ) ^{2}} - a^{p + 1} \frac{1}{p + 1} ln (a) + \frac{a ^{p + 1}}{( p + 1 ) ^{2}}

= - \frac{1}{( p + 1 ) ^{2}} - a^{p + 1} \frac{1}{p + 1} ln (a) + \frac{a ^{p + 1}}{( p + 1 ) ^{2}}

Simplify

= \frac{- 1 + a ^{p + 1}}{( 1 + p ) ^{2}} - \frac{a ^{p + 1} ln ( a )}{p + 1}

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What is the integral from a to 1 of x^pln(x) ?
The integral from a to 1 of x^pln(x) is (-1+a^{p+1})/((1+p)^2)-(a^{p+1}ln(a))/(p+1)