What is the integral from 1 to 3 of (2x-1)/((x+1)^2) ?

The integral from 1 to 3 of (2x-1)/((x+1)^2) is 2ln(2)-3/4

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integral from 1 to 3 of (2x-1)/((x+1)^2)

\int_{1}^{3} \frac{2 x - 1}{( x + 1 ) ^{2}} d x

2 ln (2) - \frac{3}{4}

Decimal

0.63629 \dots

Solution steps

\int_{1}^{3} \frac{2 x - 1}{( x + 1 ) ^{2}} d x

Expand

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

Apply the Sum Rule:

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

= \int_{1}^{3} \frac{2 x}{( x + 1 ) ^{2}} d x - \int_{1}^{3} \frac{1}{( x + 1 ) ^{2}} d x

\int_{1}^{3} \frac{2 x}{( x + 1 ) ^{2}} d x = 2 (ln (2) - \frac{1}{4})

\int_{1}^{3} \frac{1}{( x + 1 ) ^{2}} d x = \frac{1}{4}

= 2 (ln (2) - \frac{1}{4}) - \frac{1}{4}

Simplify

2 (ln (2) - \frac{1}{4}) - \frac{1}{4} : 2 ln (2) - \frac{3}{4}

= 2 ln (2) - \frac{3}{4}

What is the integral from 1 to 3 of (2x-1)/((x+1)^2) ?
The integral from 1 to 3 of (2x-1)/((x+1)^2) is 2ln(2)-3/4