What is the integral of ((3x+4)^2)/(sqrt(x)) ?

The integral of ((3x+4)^2)/(sqrt(x)) is 18/5 x^{5/2}+16x^{3/2}+32sqrt(x)+C

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integral of ((3x+4)^2)/(sqrt(x))

\int \frac{( 3 x + 4 ) ^{2}}{x} d x

\frac{18}{5} x^{\frac{5}{2}} + 16 x^{\frac{3}{2}} + 32 x + C

Solution steps

\int \frac{( 3 x + 4 ) ^{2}}{x} d x

Expand

\frac{( 3 x + 4 ) ^{2}}{x} : 9 x^{\frac{3}{2}} + 24 x + \frac{16}{x}

= \int 9 x^{\frac{3}{2}} + 24 x + \frac{16}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 9 x^{\frac{3}{2}} d x + \int 24 x d x + \int \frac{16}{x} d x

\int 9 x^{\frac{3}{2}} d x = \frac{18}{5} x^{\frac{5}{2}}

\int 24 x d x = 16 x^{\frac{3}{2}}

\int \frac{16}{x} d x = 32 x

= \frac{18}{5} x^{\frac{5}{2}} + 16 x^{\frac{3}{2}} + 32 x

Add a constant to the solution

= \frac{18}{5} x^{\frac{5}{2}} + 16 x^{\frac{3}{2}} + 32 x + C

x \to - 1 lim (x)

\frac{\partial}{\partial x} (y \cdot lo g_{e} (x))

\frac{d}{d x} (ln (7 x^{2} + 9 y^{2}))

x \to 1 lim (\frac{1}{x + 3})

\int \frac{2 v}{1 - v ^{2}} d v

What is the integral of ((3x+4)^2)/(sqrt(x)) ?
The integral of ((3x+4)^2)/(sqrt(x)) is 18/5 x^{5/2}+16x^{3/2}+32sqrt(x)+C