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

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

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

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

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

Decimal

0.88629 \dots

Solution steps

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

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

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

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

Combine the fractions

- 1 + \frac{1}{2} : - \frac{1}{2}

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

\int_{0}^{2} 2 x^{2} - 4 x d x

\int_{1}^{\infty} x^{p} d x

\int_{- 1}^{2} 2 + x - x^{2} d x

\int_{0}^{t} 2 e^{2 x} d x

\int_{0}^{2} e^{- s t} t d t

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