What is the integral of (16-16x)/(1-sqrt(x)) ?

The integral of (16-16x)/(1-sqrt(x)) is 32(1/3 x^{3/2}+x/2)+C

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integral of (16-16x)/(1-sqrt(x))

\int \frac{16 - 16 x}{1 - x} d x

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

Solution steps

\int \frac{16 - 16 x}{1 - x} d x

Apply u-substitution

= \int 32 u (u + 1) d u

Take the constant out:

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

= 32 \cdot \int u (u + 1) d u

Expand

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

= 32 \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

= 32 (\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}

= 32 (\frac{u ^{3}}{3} + \frac{u ^{2}}{2})

Substitute back

u = x

= 32 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2})

Simplify

32 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2}) : 32 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

= 32 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

Add a constant to the solution

= 32 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

\frac{d}{d x} (\frac{x ^{2} - 5 x}{x + 1})

x \to \infty lim (0. 3^{x})

\int sin (π y) d y

t an g e n t 11 x^{2} - 4 x

a re a y = 2 x^{2}, y = 16 x - 32, y = 0

What is the integral of (16-16x)/(1-sqrt(x)) ?
The integral of (16-16x)/(1-sqrt(x)) is 32(1/3 x^{3/2}+x/2)+C