What is the integral of sqrt(1+x) ?

The integral of sqrt(1+x) is 2/3 (1+x)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(1+x)

\int 1 + x d x

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

Solution steps

\int 1 + x d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + x

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

Add a constant to the solution

= \frac{2}{3} (1 + x)^{\frac{3}{2}} + C

f (x) = e^{3 x}, at x = \frac{1}{3} ln (5)

\frac{d}{d x} (sin (x) ln (2 x))

\frac{d}{d x} (x \cdot e^{- 3 x})

\int \frac{cos ( θ )}{1 + sin ^{2} ( θ )} d θ

\int (1 + y e^{x y}) d x

What is the integral of sqrt(1+x) ?
The integral of sqrt(1+x) is 2/3 (1+x)^{3/2}+C