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

The integral of (4x^3-2x^2+3)/(sqrt(x)) is 8/7 x^{7/2}-4/5 x^{5/2}+6sqrt(x)+C

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

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

\frac{8}{7} x^{\frac{7}{2}} - \frac{4}{5} x^{\frac{5}{2}} + 6 x + C

Solution steps

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

Expand

\frac{4 x ^{3} - 2 x ^{2} + 3}{x} : 4 x^{\frac{5}{2}} - 2 x^{\frac{3}{2}} + \frac{3}{x}

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

\int 4 x^{\frac{5}{2}} d x = \frac{8}{7} x^{\frac{7}{2}}

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

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

= \frac{8}{7} x^{\frac{7}{2}} - \frac{4}{5} x^{\frac{5}{2}} + 6 x

Add a constant to the solution

= \frac{8}{7} x^{\frac{7}{2}} - \frac{4}{5} x^{\frac{5}{2}} + 6 x + C

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What is the integral of (4x^3-2x^2+3)/(sqrt(x)) ?
The integral of (4x^3-2x^2+3)/(sqrt(x)) is 8/7 x^{7/2}-4/5 x^{5/2}+6sqrt(x)+C