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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

Long division

\frac{u ^{3}}{1 - u} : - u^{2} - u - 1 + \frac{1}{- u + 1}

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

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

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

\int 1 d u = u

\int \frac{1}{- u + 1} d u = - ln ∣ - u + 1 ∣

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

Substitute back

u = x

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

Simplify

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

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

Add a constant to the solution

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

\int (e^{3 x} + \frac{1}{x ^{3}} - 7 x^{\frac{3}{2}}) d x

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

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

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

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