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

The integral of (3x^2+4x)/(2sqrt(x)) is 1/15 x^{3/2}(9x+20)+C

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

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

\frac{1}{15} x^{\frac{3}{2}} (9 x + 20) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{3 x ^{2} + 4 x}{x} d x

Expand

\frac{3 x ^{2} + 4 x}{x} : \frac{3 x ^{2}}{x} + \frac{4 x}{x}

Apply the Sum Rule:

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

= \frac{1}{2} (\int \frac{3 x ^{2}}{x} d x + \int \frac{4 x}{x} d x)

\int \frac{3 x ^{2}}{x} d x = \frac{6}{5} x^{\frac{5}{2}}

\int \frac{4 x}{x} d x = \frac{8}{3} x^{\frac{3}{2}}

= \frac{1}{2} (\frac{6}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{6}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}}) : \frac{1}{15} x^{\frac{3}{2}} (9 x + 20)

= \frac{1}{15} x^{\frac{3}{2}} (9 x + 20)

Add a constant to the solution

= \frac{1}{15} x^{\frac{3}{2}} (9 x + 20) + C

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What is the integral of (3x^2+4x)/(2sqrt(x)) ?
The integral of (3x^2+4x)/(2sqrt(x)) is 1/15 x^{3/2}(9x+20)+C