What is the integral of (6-6x)/(1-sqrt(x)) ?

The integral of (6-6x)/(1-sqrt(x)) is 12(1/3 x^{3/2}+x/2)+C

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integral of (6-6x)/(1-sqrt(x))

\int \frac{6 - 6 x}{1 - x} d x

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

Solution steps

\int \frac{6 - 6 x}{1 - x} d x

Apply u-substitution

: \int 12 u (u + 1) d u

= \int 12 u (u + 1) d u

Take the constant out:

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

= 12 \cdot \int u (u + 1) d u

Expand

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

= 12 \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

= 12 (\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}

= 12 (\frac{u ^{3}}{3} + \frac{u ^{2}}{2})

Substitute back

u = x

= 12 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2})

Simplify

12 (\frac{( x ) ^{3}}{3} + \frac{( x ) ^{2}}{2}) : 12 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

= 12 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2})

Add a constant to the solution

= 12 (\frac{1}{3} x^{\frac{3}{2}} + \frac{x}{2}) + C

y = 3 x^{2} + 1, (0, - 1)

y^{^{''}} - 7 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{1}{8} x 64 - x^{2})

(5, - 4), (4, 7)

\int \frac{7}{x 36 - 49 ln ^{2} ( x )} d x

What is the integral of (6-6x)/(1-sqrt(x)) ?
The integral of (6-6x)/(1-sqrt(x)) is 12(1/3 x^{3/2}+x/2)+C