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

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

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integral of (sqrt(x)-1)

\int (x - 1) d x

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

Solution steps

\int x - 1 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 d x - \int 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

x \to 0 lim (\frac{x}{1 - x})

\frac{\partial}{\partial x} (sin (x - 3 y))

\int_{- 1}^{0} (x + 2)^{13} d x

\int_{0}^{\infty} - e^{- x} d x

f (x) = 2 x^{2} + 2 x, at x = - 2

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