What is the integral of 1/((x+1)ln^2(x+1)) ?

The integral of 1/((x+1)ln^2(x+1)) is -1/(ln(x+1))+C

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integral of 1/((x+1)ln^2(x+1))

\int \frac{1}{( x + 1 ) ln ^{2} ( x + 1 )} d x

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

Solution steps

\int \frac{1}{( x + 1 ) ln ^{2} ( x + 1 )} d x

Apply u-substitution

: \int \frac{1}{u ln ^{2} ( u )} d u

= \int \frac{1}{u ln ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

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

Add a constant to the solution

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

\int \frac{2 + 2 x}{1 - 2 x - x ^{2}} d x

\int \frac{1}{x ^{1.3}} d x

\int_{- 3}^{0} (1 + 9 - x^{2}) d x

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{4}})

n = 1 \sum \infty 4^{- 2 n}

What is the integral of 1/((x+1)ln^2(x+1)) ?
The integral of 1/((x+1)ln^2(x+1)) is -1/(ln(x+1))+C