What is the integral from 1 to e of ln^3(x) ?

The integral from 1 to e of ln^3(x) is -2e+6

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integral from 1 to e of ln^3(x)

\int_{1}^{e} ln^{3} (x) d x

- 2 e + 6

Decimal Notation

0.56343 \dots

Solution steps

\int_{1}^{e} ln^{3} (x) d x

Apply Integration By Parts

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

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

\int 3 ln^{2} (x) d x = 3 (x ln^{2} (x) - 2 (x ln (x) - x))

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

Compute the boundaries:

- 2 e + 6

= - 2 e + 6

\int_{0}^{3} x e^{x^{2}} d x

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\int_{0}^{π} 1 + sin (2 x) d x

What is the integral from 1 to e of ln^3(x) ?
The integral from 1 to e of ln^3(x) is -2e+6