What is the integral from 0 to 1 of (2x)/((x+1)^2) ?

The integral from 0 to 1 of (2x)/((x+1)^2) is 2ln(2)-1

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integral from 0 to 1 of (2x)/((x+1)^2)

\int_{0}^{1} \frac{2 x}{( x + 1 ) ^{2}} d x

2 ln (2) - 1

Decimal Notation

0.38629 \dots

Solution steps

\int_{0}^{1} \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_{0}^{1} \frac{x}{( x + 1 ) ^{2}} d x

Apply u-substitution

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

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

Expand

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

= 2 \cdot \int_{1}^{2} \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_{1}^{2} \frac{1}{u} d u - \int_{1}^{2} \frac{1}{u ^{2}} d u)

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

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

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

Simplify

2 (ln (2) - \frac{1}{2}) : 2 ln (2) - 1

= 2 ln (2) - 1

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What is the integral from 0 to 1 of (2x)/((x+1)^2) ?
The integral from 0 to 1 of (2x)/((x+1)^2) is 2ln(2)-1