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

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

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integral of (x-1)/(sqrt(x)+1)

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

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

Solution steps

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

\frac{x - 1}{x + 1} = x - 1

= \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

(- \frac{2 x}{( x ^{2} + 1 ) ^{2}})^{^{'}}

f (x) = \frac{64 x}{x ^{2} + 64}, at x = - 4

x \to 0 + lim (sin (x) n x)

\frac{d}{d x} ((2 x)^{4 x})

f (x) = x^{2} - 4 x + 3, (1, 0)

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