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

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

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integral of 1/(x(1+ln(x))^4)

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{6}{2 s + 1}

\int cot (x) ln (sin (x)) d x

\frac{d}{d x} (\frac{x - 2}{x ^{2} + 5 x + 4})

\frac{d y}{d x} = - y^{\frac{1}{3}}

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

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