What is the integral of (8-8x)/(1-sqrt(x)) ?

The integral of (8-8x)/(1-sqrt(x)) is 16(1/3 x^{3/2}+x/2)+C

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integral of (8-8x)/(1-sqrt(x))

\int \frac{8 - 8 x}{1 - x} d x

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

Solution steps

\int \frac{8 - 8 x}{1 - x} d x

Apply u-substitution

: \int 16 u (u + 1) d u

= \int 16 u (u + 1) d u

Take the constant out:

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

= 16 \cdot \int u (u + 1) d u

Expand

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

= 16 \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

= 16 (\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}

= 16 (\frac{u ^{3}}{3} + \frac{u ^{2}}{2})

Substitute back

u = x

= 16 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2})

Simplify

16 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2}) : 16 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

= 16 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

Add a constant to the solution

= 16 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

x \to 0 lim (x + \frac{1}{x})

y = ln (x^{3}) x = 12

y = (\frac{x + 2}{x - 2})^{2}, (6, 4)

\int_{0}^{\frac{π}{2}} sin (2 t) d t

\int_{0}^{4} x + 1 d x

What is the integral of (8-8x)/(1-sqrt(x)) ?
The integral of (8-8x)/(1-sqrt(x)) is 16(1/3 x^{3/2}+x/2)+C