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

4 x (x^{2} - 4)

\frac{\partial}{\partial x} (7 y - y^{3} - 4 x^{2} y)

y^{5} y^{^{'}} = - 2 cos (π x)

\int ln (3 x) d x

δ (t + 1)

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