What is the integral of (18-18x)/(1-sqrt(x)) ?

The integral of (18-18x)/(1-sqrt(x)) is 36(1/3 x^{3/2}+x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (18-18x)/(1-sqrt(x))

\int \frac{18 - 18 x}{1 - x} d x

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

Solution steps

\int \frac{18 - 18 x}{1 - x} d x

Apply u-substitution

= \int 36 u (u + 1) d u

Take the constant out:

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

= 36 \cdot \int u (u + 1) d u

Expand

u (u + 1) : u^{2} + u

= 36 \cdot \int u^{2} + u 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

= 36 (\int u^{2} d u + \int u d u)

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

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

= 36 (\frac{u ^{3}}{3} + \frac{u ^{2}}{2})

Substitute back

u = x

= 36 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2})

Simplify

36 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2}) : 36 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

= 36 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

Add a constant to the solution

= 36 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

x \to 0 + lim ((5 x)^{2 x})

f (x) = x^{2} + x + 4

\frac{\partial}{\partial u} (2 u + v)

\int 5 x^{3} 3 (2 x^{4} - 4)^{2} d x

x \to 0 lim (\frac{cos ( x )}{96})

What is the integral of (18-18x)/(1-sqrt(x)) ?
The integral of (18-18x)/(1-sqrt(x)) is 36(1/3 x^{3/2}+x/2)+C