What is the integral of 1/(x(ln(x))^9) ?

The integral of 1/(x(ln(x))^9) is -1/(8ln^8(x))+C

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integral of 1/(x(ln(x))^9)

\int \frac{1}{x ( ln ( x ) ) ^{9}} d x

- \frac{1}{8 ln ^{8} ( x )} + C

Solution steps

\int \frac{1}{x ( ln ( x ) ) ^{9}} d x

Simplify

= \int \frac{1}{x ln ^{9} ( x )} d x

Apply u-substitution

: \int \frac{1}{u ^{9}} d u

= \int \frac{1}{u ^{9}} d u

Apply the Power Rule

: - \frac{1}{8 u ^{8}}

= - \frac{1}{8 u ^{8}}

Substitute back

u = ln (x)

= - \frac{1}{8 ln ^{8} ( x )}

Add a constant to the solution

= - \frac{1}{8 ln ^{8} ( x )} + C

\frac{\partial}{\partial x} (x^{2} - 2 x y + 3 y^{2} - 8 y)

\frac{d}{d x} ((f (x))^{3})

\frac{\partial}{\partial x} (ln (9 x^{2} + y^{2} + 9))

x \to 4 lim (x^{2} - 6 x + 8)

n = 1 \sum \infty (- 1)^{n} \cdot n

What is the integral of 1/(x(ln(x))^9) ?
The integral of 1/(x(ln(x))^9) is -1/(8ln^8(x))+C