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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

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

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

Compute the boundaries:

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

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

Simplify

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

\frac{d ^{2}}{d x ^{2}} (e^{6 e^{x}})

f (x) = x^{4} - 4 x^{3} - 20 x^{2} + 96 x + 60

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 2

\int cos^{2} (x s i) n^{2} x d x

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

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