What is the integral of (4x^2-6x+3)/(sqrt(x)) ?

The integral of (4x^2-6x+3)/(sqrt(x)) is 8/5 x^{5/2}-4x^{3/2}+6sqrt(x)+C

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integral of (4x^2-6x+3)/(sqrt(x))

\int \frac{4 x ^{2} - 6 x + 3}{x} d x

\frac{8}{5} x^{\frac{5}{2}} - 4 x^{\frac{3}{2}} + 6 x + C

Solution steps

\int \frac{4 x ^{2} - 6 x + 3}{x} d x

Expand

\frac{4 x ^{2} - 6 x + 3}{x} : 4 x^{\frac{3}{2}} - 6 x + \frac{3}{x}

= \int 4 x^{\frac{3}{2}} - 6 x + \frac{3}{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 4 x^{\frac{3}{2}} d x - \int 6 x d x + \int \frac{3}{x} d x

\int 4 x^{\frac{3}{2}} d x = \frac{8}{5} x^{\frac{5}{2}}

\int 6 x d x = 4 x^{\frac{3}{2}}

\int \frac{3}{x} d x = 6 x

= \frac{8}{5} x^{\frac{5}{2}} - 4 x^{\frac{3}{2}} + 6 x

Add a constant to the solution

= \frac{8}{5} x^{\frac{5}{2}} - 4 x^{\frac{3}{2}} + 6 x + C

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

\frac{d}{d x} (tan (x) - cot (x))

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

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

x \to + 0 lim (\frac{14}{tan ( x )})

What is the integral of (4x^2-6x+3)/(sqrt(x)) ?
The integral of (4x^2-6x+3)/(sqrt(x)) is 8/5 x^{5/2}-4x^{3/2}+6sqrt(x)+C