What is the integral of 1/(x(ln(x))^8) ?

The integral of 1/(x(ln(x))^8) is -1/(7ln^7(x))+C

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integral of 1/(x(ln(x))^8)

\int \frac{1}{x ( ln ( x ) ) ^{8}} d x

- \frac{1}{7 ln ^{7} ( x )} + C

Solution steps

\int \frac{1}{x ( ln ( x ) ) ^{8}} d x

Apply u-substitution

: \int \frac{1}{u ^{8}} d u

= \int \frac{1}{u ^{8}} d u

Apply the Power Rule

: - \frac{1}{7 u ^{7}}

= - \frac{1}{7 u ^{7}}

Substitute back

u = ln (x)

= - \frac{1}{7 ln ^{7} ( x )}

Add a constant to the solution

= - \frac{1}{7 ln ^{7} ( x )} + C

\int \frac{2 + x + x}{x} d x

(sec (x))^{^{'}}

0.334 \cdot sin (0.3494 x - 0.06988)

\frac{d}{d x} (6 + sec (π) x^{2})

sin (x), [0, π]

What is the integral of 1/(x(ln(x))^8) ?
The integral of 1/(x(ln(x))^8) is -1/(7ln^7(x))+C