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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((sqrt(x)+1))

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

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

Solution steps

\int \frac{x}{x + 1} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (\frac{v ^{3}}{3} - \frac{3 v ^{2}}{2} + 3 v - ln ∣ v ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{1 - x}{9 - 4 x} d x

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

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

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

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