What is the integral of x/(sqrt(3x+4)) ?

The integral of x/(sqrt(3x+4)) is 2/9 (1/3 (3x+4)^{3/2}-4sqrt(3x+4))+C

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integral of x/(sqrt(3x+4))

\int \frac{x}{3 x + 4} d x

\frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4) + C

Solution steps

\int \frac{x}{3 x + 4} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} - 4 )}{9} d u

= \int \frac{2 ( u ^{2} - 4 )}{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} - 4 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 4 d u)

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

\int 4 d u = 4 u

= \frac{2}{9} (\frac{u ^{3}}{3} - 4 u)

Substitute back

u = 3 x + 4

= \frac{2}{9} (\frac{( 3 x + 4 ) ^{3}}{3} - 4 3 x + 4)

Simplify

\frac{2}{9} (\frac{( 3 x + 4 ) ^{3}}{3} - 4 3 x + 4) : \frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4)

= \frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4)

Add a constant to the solution

= \frac{2}{9} (\frac{1}{3} (3 x + 4)^{\frac{3}{2}} - 4 3 x + 4) + C

\frac{\partial}{\partial x} (2 x^{2} + y)

\frac{\partial}{\partial x} (\frac{1}{x})

\frac{d}{d x} (\frac{sin ^{2} ( x )}{x ^{2}})

\frac{d}{d x} (3^{3 x})

\frac{\partial}{\partial x} (2 e^{x^{9} y})

What is the integral of x/(sqrt(3x+4)) ?
The integral of x/(sqrt(3x+4)) is 2/9 (1/3 (3x+4)^{3/2}-4sqrt(3x+4))+C