What is the integral of x/(sqrt(36+3x)) ?

The integral of x/(sqrt(36+3x)) is 2/9 (1/3 (36+3x)^{3/2}-36sqrt(36+3x))+C

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integral of x/(sqrt(36+3x))

\int \frac{x}{36 + 3 x} d x

\frac{2}{9} (\frac{1}{3} (36 + 3 x)^{\frac{3}{2}} - 36 36 + 3 x) + C

Solution steps

\int \frac{x}{36 + 3 x} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} - 36 )}{9} d u

= \int \frac{2 ( u ^{2} - 36 )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int u^{2} - 36 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

= \frac{2}{9} (\int u^{2} d u - \int 36 d u)

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

\int 36 d u = 36 u

= \frac{2}{9} (\frac{u ^{3}}{3} - 36 u)

Substitute back

u = 36 + 3 x

= \frac{2}{9} (\frac{( 36 + 3 x ) ^{3}}{3} - 36 36 + 3 x)

Simplify

\frac{2}{9} (\frac{( 36 + 3 x ) ^{3}}{3} - 36 36 + 3 x) : \frac{2}{9} (\frac{1}{3} (36 + 3 x)^{\frac{3}{2}} - 36 36 + 3 x)

= \frac{2}{9} (\frac{1}{3} (36 + 3 x)^{\frac{3}{2}} - 36 36 + 3 x)

Add a constant to the solution

= \frac{2}{9} (\frac{1}{3} (36 + 3 x)^{\frac{3}{2}} - 36 36 + 3 x) + C

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What is the integral of x/(sqrt(36+3x)) ?
The integral of x/(sqrt(36+3x)) is 2/9 (1/3 (36+3x)^{3/2}-36sqrt(36+3x))+C