What is the integral of 4/((x+1)^2) ?

The integral of 4/((x+1)^2) is -4/(x+1)+C

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integral of 4/((x+1)^2)

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

- \frac{4}{x + 1} + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{1}{( x + 1 ) ^{2}} d x

Apply u-substitution

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

= 4 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 4 (- \frac{1}{u})

Substitute back

u = x + 1

= 4 (- \frac{1}{x + 1})

Simplify

4 (- \frac{1}{x + 1}) : - \frac{4}{x + 1}

= - \frac{4}{x + 1}

Add a constant to the solution

= - \frac{4}{x + 1} + C

\frac{d}{d x} (16 sin (2 x))

\frac{1}{2 s ^{2} + 4 s + 3}

y^{^{'}} + 4 y = 3 sin (e^{4 x})

\int \frac{4 x ^{2} - 16}{x} d x

\frac{d}{d x} ((tan (x))^{- 1})