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

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

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

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

1 - ln (2)

Decimal

0.30685 \dots

Solution steps

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

Apply u-substitution

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

Expand

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

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

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

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

= 1 - ln (2)

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

\int_{0}^{1} \frac{2}{3} (x + 2 y) d x

\frac{d}{d x} (a x + cot (a x))

x \to - 4 + lim (\frac{- 4 x}{x + 4})

f (x) = 2 sin^{2} (x)

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