What is the integral from 1 to e of (ln(x))^5 ?

The integral from 1 to e of (ln(x))^5 is -44e+120

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integral from 1 to e of (ln(x))^5

\int_{1}^{e} (ln (x))^{5} d x

- 44 e + 120

Decimal Notation

0.39559 \dots

Solution steps

\int_{1}^{e} (ln (x))^{5} d x

Simplify

= \int_{1}^{e} ln^{5} (x) d x

Apply Integration By Parts

: [x ln^{5} (x) - \int 5 ln^{4} (x) d x]_{1}^{e}

= [x ln^{5} (x) - \int 5 ln^{4} (x) d x]_{1}^{e}

\int 5 ln^{4} (x) d x = 5 (x ln^{4} (x) - 4 (x ln^{3} (x) - 3 (x ln^{2} (x) - 2 (x ln (x) - x))))

= [x ln^{5} (x) - 5 (x ln^{4} (x) - 4 (x ln^{3} (x) - 3 (x ln^{2} (x) - 2 (x ln (x) - x))))]_{1}^{e}

Compute the boundaries:

- 44 e + 120

= - 44 e + 120

\int_{0}^{5} \frac{1}{( x - 1 ) ^{4}} d x

\int_{0}^{4} (3 x + 4)^{\frac{3}{2}} d x

\int_{1}^{e} \frac{( ln ( x ^{2} ) ) ^{2}}{x} d x

\int_{2}^{5} (- 2 x^{2} + 12 x - 17) d x

\int_{5}^{6} x (x - 5)^{12} d x

What is the integral from 1 to e of (ln(x))^5 ?
The integral from 1 to e of (ln(x))^5 is -44e+120