What is the integral of 1/((1+sqrt(x))^3) ?

The integral of 1/((1+sqrt(x))^3) is -2/(1+sqrt(x))+1/((1+sqrt(x))^2)+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(x))^3)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v - 1}{v ^{3}} d v

= 2 \cdot \int \frac{v - 1}{v ^{3}} d v

Expand

\frac{v - 1}{v ^{3}} : \frac{1}{v ^{2}} - \frac{1}{v ^{3}}

= 2 \cdot \int \frac{1}{v ^{2}} - \frac{1}{v ^{3}} d v

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 \frac{1}{v ^{2}} d v - \int \frac{1}{v ^{3}} d v)

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

= 2 (- \frac{1}{v} - (- \frac{1}{2 v ^{2}}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \to \infty lim (2 x^{3})

\int x^{2} sin (8 x) d x

arcsin (3 x) + arccos (3 x)

\frac{d}{d x} (- \frac{2 x ( - x ^{2} + 27 )}{( x ^{2} + 9 ) ^{3}})

x \to 0 + lim (x^{10 x})

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