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

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

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

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

- \frac{2}{x + 1} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= 2 (- \frac{1}{u})

Substitute back

u = x + 1

= 2 (- \frac{1}{x + 1})

Simplify

2 (- \frac{1}{x + 1}) : - \frac{2}{x + 1}

= - \frac{2}{x + 1}

Add a constant to the solution

= - \frac{2}{x + 1} + C

\int 1 + e^{x} y + x e^{x} y d x

x \to 1 lim (3 x^{2} + 1)

\int (2 - y) d y

\frac{d ^{2}}{d x ^{2}} (6 ln (4 x))

x \to - 1 - lim (\frac{1}{x ^{2} + 1})