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

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

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

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

2 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) + C

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 2 (ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = x + 1

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

Simplify

= 2 (ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Add a constant to the solution

= 2 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) + C

g (t) = t^{3} + 3 t

5 e^{x} cos (x)

\frac{d}{d x} (arctan (\frac{300}{x}))

h (x) = ln (x + x^{2} - 6)

\int sec^{6} (6 x) d x

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