What is the integral of x/(sqrt(25+3x)) ?

The integral of x/(sqrt(25+3x)) is 2/9 (1/3 (25+3x)^{3/2}-25sqrt(25+3x))+C

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integral of x/(sqrt(25+3x))

\int \frac{x}{25 + 3 x} d x

\frac{2}{9} (\frac{1}{3} (25 + 3 x)^{\frac{3}{2}} - 25 25 + 3 x) + C

Solution steps

\int \frac{x}{25 + 3 x} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} - 25 )}{9} d u

= \int \frac{2 ( u ^{2} - 25 )}{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} - 25 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 25 d u)

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

\int 25 d u = 25 u

= \frac{2}{9} (\frac{u ^{3}}{3} - 25 u)

Substitute back

u = 25 + 3 x

= \frac{2}{9} (\frac{( 25 + 3 x ) ^{3}}{3} - 25 25 + 3 x)

Simplify

\frac{2}{9} (\frac{( 25 + 3 x ) ^{3}}{3} - 25 25 + 3 x) : \frac{2}{9} (\frac{1}{3} (25 + 3 x)^{\frac{3}{2}} - 25 25 + 3 x)

= \frac{2}{9} (\frac{1}{3} (25 + 3 x)^{\frac{3}{2}} - 25 25 + 3 x)

Add a constant to the solution

= \frac{2}{9} (\frac{1}{3} (25 + 3 x)^{\frac{3}{2}} - 25 25 + 3 x) + C

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What is the integral of x/(sqrt(25+3x)) ?
The integral of x/(sqrt(25+3x)) is 2/9 (1/3 (25+3x)^{3/2}-25sqrt(25+3x))+C