What is the integral from x^2 to e^x of t^2ln(t) ?

The integral from x^2 to e^x of t^2ln(t) is (xe^{3x}-2x^6ln(x))/3+(-e^{3x}+x^6)/9

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integral from x^2 to e^x of t^2ln(t)

\int_{x^{2}}^{e^{x}} t^{2} ln (t) d t

\frac{x e ^{3 x} - 2 x ^{6} ln ( x )}{3} + \frac{- e ^{3 x} + x ^{6}}{9}

Solution steps

\int_{x^{2}}^{e^{x}} t^{2} ln (t) d t

Apply Integration By Parts

: [\frac{1}{3} t^{3} ln (t) - \int \frac{t ^{2}}{3} d t]_{x^{2}}^{e^{x}}

= [\frac{1}{3} t^{3} ln (t) - \int \frac{t ^{2}}{3} d t]_{x^{2}}^{e^{x}}

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

= [\frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9}]_{x^{2}}^{e^{x}}

Compute the boundaries:

\frac{1}{3} x e^{3 x} - \frac{e ^{3 x}}{9} - \frac{2}{3} x^{6} ln (x) + \frac{x ^{6}}{9}

= \frac{1}{3} x e^{3 x} - \frac{e ^{3 x}}{9} - \frac{2}{3} x^{6} ln (x) + \frac{x ^{6}}{9}

Simplify

= \frac{x e ^{3 x} - 2 x ^{6} ln ( x )}{3} + \frac{- e ^{3 x} + x ^{6}}{9}

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What is the integral from x^2 to e^x of t^2ln(t) ?
The integral from x^2 to e^x of t^2ln(t) is (xe^{3x}-2x^6ln(x))/3+(-e^{3x}+x^6)/9