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

The integral of 2xsqrt(1+x) is 2(2/5 (1+x)^{5/2}-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 2xsqrt(1+x)

\int 2 x 1 + x d x

2 (\frac{2}{5} (1 + x)^{\frac{5}{2}} - \frac{2}{3} (1 + x)^{\frac{3}{2}}) + C

Solution steps

\int 2 x 1 + x 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 + x d x

Apply u-substitution

: \int (u - 1) u d u

= 2 \cdot \int (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

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

= 2 (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= 2 (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + x

= 2 (\frac{2}{5} (1 + x)^{\frac{5}{2}} - \frac{2}{3} (1 + x)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{2}{5} (1 + x)^{\frac{5}{2}} - \frac{2}{3} (1 + x)^{\frac{3}{2}}) + C

\int e^{x} cos (n) x d x

\int (\frac{5}{1 - x ^{2}} - \frac{2}{x ^{2} + 1}) d x

\int x^{2} r^{2} - x^{2} d x

\int x^{2} - 3 y^{2} d y

\int tan (x) \cdot (sec (x))^{2} d x

What is the integral of 2xsqrt(1+x) ?
The integral of 2xsqrt(1+x) is 2(2/5 (1+x)^{5/2}-2/3 (1+x)^{3/2})+C