What is the integral from 1 to e of (ln(x))^7 ?

The integral from 1 to e of (ln(x))^7 is -1854e+5040

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integral from 1 to e of (ln(x))^7

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

- 1854 e + 5040

Decimal Notation

0.30549 \dots

Solution steps

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

Simplify

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

Apply u-substitution

: \int_{0}^{1} u^{7} e^{u} d u

= \int_{0}^{1} u^{7} e^{u} d u

Apply Integration By Parts

: [u^{7} e^{u} - \int 7 u^{6} e^{u} d u]_{0}^{1}

= [u^{7} e^{u} - \int 7 u^{6} e^{u} d u]_{0}^{1}

\int 7 u^{6} e^{u} d u = 7 (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}))))))

= [u^{7} e^{u} - 7 (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:

- 1854 e + 5040

= - 1854 e + 5040

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

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

\int_{0}^{6} 4 + 2 x d x

\int_{0}^{1} (1 - x^{4})^{3} d x

\int_{2}^{\infty} ln (2 x - 3) d x

What is the integral from 1 to e of (ln(x))^7 ?
The integral from 1 to e of (ln(x))^7 is -1854e+5040