What is the integral from 1 to e of (ln(x))^6 ?

The integral from 1 to e of (ln(x))^6 is 265e-720

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integral from 1 to e of (ln(x))^6

\int_{1}^{e} (ln (x))^{6} d x

265 e - 720

Decimal Notation

0.34468 \dots

Solution steps

\int_{1}^{e} (ln (x))^{6} d x

Simplify

= \int_{1}^{e} ln^{6} (x) d x

Apply u-substitution

: \int_{0}^{1} u^{6} e^{u} d u

= \int_{0}^{1} u^{6} e^{u} d u

Apply Integration By Parts

: [u^{6} e^{u} - \int 6 u^{5} e^{u} d u]_{0}^{1}

= [u^{6} e^{u} - \int 6 u^{5} e^{u} d u]_{0}^{1}

\int 6 u^{5} e^{u} d u = 6 (u^{5} e^{u} - 5 (u^{4} e^{u} - 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))))

= [u^{6} e^{u} - 6 (u^{5} e^{u} - 5 (u^{4} e^{u} - 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))))]_{0}^{1}

Compute the boundaries:

265 e - 720

= 265 e - 720

\int_{0}^{a} \frac{1}{2 x + 1} d x

\int_{0}^{π} sin (θ) d θ

\int_{1}^{e} (ln (x))^{7} d x

\int_{0}^{1} x^{6} e^{- x} d x

\int_{0}^{1} x^{7} e^{- x} d x

What is the integral from 1 to e of (ln(x))^6 ?
The integral from 1 to e of (ln(x))^6 is 265e-720