What is the integral from 1 to e of x^nln(x) ?

The integral from 1 to e of x^nln(x) is (ne^{n+1}+1)/((1+n)^2)

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integral from 1 to e of x^nln(x)

\int_{1}^{e} x^{n} ln (x) d x

\frac{n e ^{n + 1} + 1}{( 1 + n ) ^{2}}

Solution steps

\int_{1}^{e} x^{n} ln (x) d x

Apply Integration By Parts

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

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

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

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

\frac{x ^{n + 1}}{( n + 1 ) ^{2}} (n + 1) = \frac{x ^{n + 1}}{n + 1}

= [\frac{1}{n + 1} (x^{n + 1} ln (x) - \frac{x ^{n + 1}}{n + 1})]_{1}^{e}

Compute the boundaries:

\frac{n e ^{n + 1}}{( n + 1 ) ^{2}} + \frac{1}{( n + 1 ) ^{2}}

= \frac{n e ^{n + 1}}{( n + 1 ) ^{2}} + \frac{1}{( n + 1 ) ^{2}}

Simplify

= \frac{n e ^{n + 1} + 1}{( 1 + n ) ^{2}}

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What is the integral from 1 to e of x^nln(x) ?
The integral from 1 to e of x^nln(x) is (ne^{n+1}+1)/((1+n)^2)