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

The integral from 0 to 1 of x/(sqrt(1+x)) is 2((2sqrt(2)-1)/3-sqrt(2)+1)

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

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

2 (\frac{2 2 - 1}{3} - 2 + 1)

Decimal Notation

0.39052 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} 2 (u^{2} - 1) d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

\int_{1}^{2} u^{2} d u = \frac{2 2 - 1}{3}

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

= 2 (\frac{2 2 - 1}{3} - (2 - 1))

Simplify

= 2 (\frac{2 2 - 1}{3} - 2 + 1)

\frac{d}{d t} (e^{3 t})

\frac{d}{d x} (a cos (t) + b sin (t))

y = 6 x, y = \frac{2}{3} x, y = 40 - x^{2}

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

\frac{\partial}{\partial x} (x^{2} \cdot cos (x y))

What is the integral from 0 to 1 of x/(sqrt(1+x)) ?
The integral from 0 to 1 of x/(sqrt(1+x)) is 2((2sqrt(2)-1)/3-sqrt(2)+1)