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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((1+sqrt(1+x))^{1/2)}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= 2 \cdot \int \frac{u - 1}{u ^{\frac{1}{2}}} d u

Expand

\frac{u - 1}{u ^{\frac{1}{2}}} : u^{\frac{1}{2}} - \frac{1}{u ^{\frac{1}{2}}}

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

\int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}}

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

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

Substitute back

u = 1 + 1 + x

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

Apply radical rule:

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

Add a constant to the solution

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

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

\int (4 y - x^{2})^{2} d x

\int x^{3} 1 + 100 x^{2} d x

\int \frac{x ^{2}}{25 - 16 x ^{2}} d x

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

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