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

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

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

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

- \frac{1}{x + 1} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 1

= - \frac{1}{x + 1}

Add a constant to the solution

= - \frac{1}{x + 1} + C

\frac{\partial}{\partial y} (x e^{yz})

\int \frac{1}{4 sec ( y )} d y

\frac{d y}{d x} = \frac{x x ^{2} - 1}{y}

\int (3 p^{2} - 2)^{2} e^{- 4 p} d p

2 x^{2} + 2 x + 3

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