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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

= \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}}}

= \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

= \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}}

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

\int \frac{1}{4 sin ^{2} ( x ) + cos ^{2} ( x )} d x

\int \frac{3 x + 23}{x ^{2} - 8 x - 9} d x

\int e^{8 x - 6} d x

\int \frac{( 4 + ln ( x ) ) ^{5}}{4 x} d x

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