What is the integral of 2sqrt(x+1) ?

The integral of 2sqrt(x+1) is 4/3 (x+1)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2sqrt(x+1)

\int 2 x + 1 d x

\frac{4}{3} (x + 1)^{\frac{3}{2}} + C

Solution steps

\int 2 x + 1 d x

Take the constant out:

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

= 2 \cdot \int x + 1 d x

Apply u-substitution

= 2 \cdot \int u d u

Apply the Power Rule

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

Substitute back

u = x + 1

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

Simplify

2 \cdot \frac{2}{3} (x + 1)^{\frac{3}{2}} : \frac{4}{3} (x + 1)^{\frac{3}{2}}

= \frac{4}{3} (x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{4}{3} (x + 1)^{\frac{3}{2}} + C

What is the integral of 2sqrt(x+1) ?
The integral of 2sqrt(x+1) is 4/3 (x+1)^{3/2}+C