What is the integral of (20-20x)/(1-sqrt(x)) ?

The integral of (20-20x)/(1-sqrt(x)) is 40(1/3 x^{3/2}+x/2)+C

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integral of (20-20x)/(1-sqrt(x))

\int \frac{20 - 20 x}{1 - x} d x

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

Solution steps

\int \frac{20 - 20 x}{1 - x} d x

Apply u-substitution

: \int 40 u (u + 1) d u

= \int 40 u (u + 1) d u

Take the constant out:

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

= 40 \cdot \int u (u + 1) d u

Expand

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

= 40 \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

= 40 (\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}

= 40 (\frac{u ^{3}}{3} + \frac{u ^{2}}{2})

Substitute back

u = x

= 40 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2})

Simplify

40 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2}) : 40 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

= 40 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

Add a constant to the solution

= 40 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

\int ln (5 - x) d x

f (x) = 2 x^{2} - 7 x + 5

\int \frac{x ^{3} + 2 x ^{2} - 18 x + 2}{x ^{2} + x - 12} d x

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 7

x \to 1 lim (- 2 x + 3)

What is the integral of (20-20x)/(1-sqrt(x)) ?
The integral of (20-20x)/(1-sqrt(x)) is 40(1/3 x^{3/2}+x/2)+C