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

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

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

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

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

Solution steps

\int \frac{x}{x + 9} 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 = x + 9

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

Simplify

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

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

Add a constant to the solution

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

\int (3^{(2 - x)} + (x + 1)^{2} e^{(1 - x)}) d x

y = 3 x^{2} - 6, (- 1, - 3)

\int_{2}^{\infty} e^{- 9 p} d p

\int 5^{x^{2} + 2 x + 3} (x + 1) d x

\int 5 x^{\frac{1}{4}} d x

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