What is the integral of x/(sqrt(9-x)) ?

The integral of x/(sqrt(9-x)) is -2(-1/3 (9-x)^{3/2}+9sqrt(9-x))+C

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integral of x/(sqrt(9-x))

\int \frac{x}{9 - x} d x

- 2 (- \frac{1}{3} (9 - x)^{\frac{3}{2}} + 9 9 - x) + C

Solution steps

\int \frac{x}{9 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 9) d u

= \int - 2 (- u^{2} + 9) d u

Take the constant out:

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

= - 2 \cdot \int - u^{2} + 9 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

= - 2 (- \int u^{2} d u + \int 9 d u)

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

\int 9 d u = 9 u

= - 2 (- \frac{u ^{3}}{3} + 9 u)

Substitute back

u = 9 - x

= - 2 (- \frac{( 9 - x ) ^{3}}{3} + 9 9 - x)

Simplify

- 2 (- \frac{( 9 - x ) ^{3}}{3} + 9 9 - x) : - 2 (- \frac{1}{3} (9 - x)^{\frac{3}{2}} + 9 9 - x)

= - 2 (- \frac{1}{3} (9 - x)^{\frac{3}{2}} + 9 9 - x)

Add a constant to the solution

= - 2 (- \frac{1}{3} (9 - x)^{\frac{3}{2}} + 9 9 - x) + C

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

\int 1 - 9 t^{2} d t

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What is the integral of x/(sqrt(9-x)) ?
The integral of x/(sqrt(9-x)) is -2(-1/3 (9-x)^{3/2}+9sqrt(9-x))+C