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

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

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

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

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

Solution steps

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

Take the constant out:

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

= k \cdot \int \frac{x}{1 - x} d x

Apply u-substitution

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

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

Take the constant out:

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

= k (- 2 \cdot \int - 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

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

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

\int 1 d u = u

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

Substitute back

u = 1 - x

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

Simplify

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

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

Add a constant to the solution

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

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