What is the integral of (-3x^2)/(sqrt(1-x)) ?

The integral of (-3x^2)/(sqrt(1-x)) is 6(1/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+sqrt(1-x))+C

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integral of (-3x^2)/(sqrt(1-x))

\int \frac{- 3 x ^{2}}{1 - x} d x

6 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x) + C

Solution steps

\int \frac{- 3 x ^{2}}{1 - x} d x

Take the constant out:

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

= - 3 \cdot \int \frac{x ^{2}}{1 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 1)^{2} d u

= - 3 \cdot \int - 2 (- u^{2} + 1)^{2} d u

Take the constant out:

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

= - 3 (- 2 \cdot \int (- u^{2} + 1)^{2} d u)

Expand

(- u^{2} + 1)^{2} : u^{4} - 2 u^{2} + 1

= - 3 (- 2 \cdot \int u^{4} - 2 u^{2} + 1 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

= - 3 (- 2 (\int u^{4} d u - \int 2 u^{2} d u + \int 1 d u))

\int u^{4} d u = \frac{u ^{5}}{5}

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

\int 1 d u = u

= - 3 (- 2 (\frac{u ^{5}}{5} - \frac{2 u ^{3}}{3} + u))

Substitute back

u = 1 - x

= - 3 (- 2 (\frac{( 1 - x ) ^{5}}{5} - \frac{2 ( 1 - x ) ^{3}}{3} + 1 - x))

Simplify

- 3 (- 2 (\frac{( 1 - x ) ^{5}}{5} - \frac{2 ( 1 - x ) ^{3}}{3} + 1 - x)) : 6 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

= 6 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

Add a constant to the solution

= 6 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x) + C

\int 8 ln (x^{2} - 1) d x

\frac{\partial}{\partial y} (z arctan (\frac{y}{x}))

f (x) = \frac{1}{x ln ( x )}

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

tan (3 t)

What is the integral of (-3x^2)/(sqrt(1-x)) ?
The integral of (-3x^2)/(sqrt(1-x)) is 6(1/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+sqrt(1-x))+C