What is the integral of (x^2-2x-1)/(sqrt(x)) ?

The integral of (x^2-2x-1)/(sqrt(x)) is 2/5 x^{5/2}-4/3 x^{3/2}-2sqrt(x)+C

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integral of (x^2-2x-1)/(sqrt(x))

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

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

Solution steps

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

Expand

\frac{x ^{2} - 2 x - 1}{x} : x^{\frac{3}{2}} - 2 x - \frac{1}{x}

= \int x^{\frac{3}{2}} - 2 x - \frac{1}{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 2 x d x - \int \frac{1}{x} d x

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

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

\int \frac{1}{x} d x = 2 x

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

Add a constant to the solution

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

e^{1 - t}

x y^{^{'}} sin^{2} (\frac{y}{x}) = x + y sin^{2} (\frac{y}{x})

\frac{\partial}{\partial x} (8 y^{7} cos (8 x))

\frac{1}{b ^{\frac{2}{7}}}

x \to 2 lim (\frac{3}{x ^{2} - 4})

What is the integral of (x^2-2x-1)/(sqrt(x)) ?
The integral of (x^2-2x-1)/(sqrt(x)) is 2/5 x^{5/2}-4/3 x^{3/2}-2sqrt(x)+C