What is the integral of (6x^2-5x^4-2)/(sqrt(x)) ?

The integral of (6x^2-5x^4-2)/(sqrt(x)) is 12/5 x^{5/2}-10/9 x^{9/2}-4sqrt(x)+C

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integral of (6x^2-5x^4-2)/(sqrt(x))

\int \frac{6 x ^{2} - 5 x ^{4} - 2}{x} d x

\frac{12}{5} x^{\frac{5}{2}} - \frac{10}{9} x^{\frac{9}{2}} - 4 x + C

Solution steps

\int \frac{6 x ^{2} - 5 x ^{4} - 2}{x} d x

Expand

\frac{6 x ^{2} - 5 x ^{4} - 2}{x} : 6 x^{\frac{3}{2}} - 5 x^{\frac{7}{2}} - \frac{2}{x}

= \int 6 x^{\frac{3}{2}} - 5 x^{\frac{7}{2}} - \frac{2}{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 6 x^{\frac{3}{2}} d x - \int 5 x^{\frac{7}{2}} d x - \int \frac{2}{x} d x

\int 6 x^{\frac{3}{2}} d x = \frac{12}{5} x^{\frac{5}{2}}

\int 5 x^{\frac{7}{2}} d x = \frac{10}{9} x^{\frac{9}{2}}

\int \frac{2}{x} d x = 4 x

= \frac{12}{5} x^{\frac{5}{2}} - \frac{10}{9} x^{\frac{9}{2}} - 4 x

Add a constant to the solution

= \frac{12}{5} x^{\frac{5}{2}} - \frac{10}{9} x^{\frac{9}{2}} - 4 x + C

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What is the integral of (6x^2-5x^4-2)/(sqrt(x)) ?
The integral of (6x^2-5x^4-2)/(sqrt(x)) is 12/5 x^{5/2}-10/9 x^{9/2}-4sqrt(x)+C