What is the integral of 1/(xln^4(x)) ?

The integral of 1/(xln^4(x)) is -1/(3ln^3(x))+C

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integral of 1/(xln^4(x))

\int \frac{1}{x ln ^{4} ( x )} d x

- \frac{1}{3 ln ^{3} ( x )} + C

Solution steps

\int \frac{1}{x ln ^{4} ( x )} d x

Apply u-substitution

: \int \frac{1}{u ^{4}} d u

= \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= - \frac{1}{3 u ^{3}}

Substitute back

u = ln (x)

= - \frac{1}{3 ln ^{3} ( x )}

Add a constant to the solution

= - \frac{1}{3 ln ^{3} ( x )} + C

y^{^{'}} = y^{\frac{1}{3}}

\frac{d y}{d x} = 5 x + 15 y

f (x) = \frac{x}{4 x + 3 ln ( x )}

y = \frac{- 4 x ^{2} + 12}{2 x + 2}

\frac{d}{d x} ((2 x - 3 x^{3}) tan (x))

What is the integral of 1/(xln^4(x)) ?
The integral of 1/(xln^4(x)) is -1/(3ln^3(x))+C