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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

\int \frac{4}{x} d x = 8 x

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

Add a constant to the solution

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

\int (1 - x)^{- \frac{1}{2}} d x

\int \frac{1}{x + x} d x

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

\int \frac{1}{( 2 x + 1 ) ( x + 2 )} d x

\int \frac{5 x ^{2} - x - 4}{x ^{2} ( x + 1 )} d x

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