What is the integral from 0 to+oo of x*(x+1)^{-2} ?

The integral from 0 to+oo of x*(x+1)^{-2} is ln|o^2+1|+1/(o^2+1)-1

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integral from 0 to+oo of x*(x+1)^{-2}

\int_{0}^{+ oo} x \cdot (x + 1)^{- 2} d x

ln o^{2} + 1 + \frac{1}{o ^{2} + 1} - 1

Solution steps

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

Simplify

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

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

Apply u-substitution

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

Expand

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

= \int_{1}^{o^{2} + 1} \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

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

\int_{1}^{o^{2} + 1} \frac{1}{u} d u = ln o^{2} + 1

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

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

Simplify

= ln o^{2} + 1 + \frac{1}{o ^{2} + 1} - 1

f (x) = 5 - x

\int_{0}^{3} 1 + x d x

\frac{d}{d x} (\frac{3}{5 x})

\frac{d}{d t} (2 sin (t) + 2)

\frac{d}{d x} (e^{0.2 x})

What is the integral from 0 to+oo of x*(x+1)^{-2} ?
The integral from 0 to+oo of x*(x+1)^{-2} is ln|o^2+1|+1/(o^2+1)-1