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

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

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

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

\frac{2}{3} (x + 1)^{\frac{3}{2}} - x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{u + 1} d u

= \int \frac{u - 1}{u + 1} d u

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

= \int u - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u - \int 1 d u

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

\int 1 d u = u

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

Substitute back

u = x + 1

= \frac{2}{3} (x + 1)^{\frac{3}{2}} - (x + 1)

Simplify

= \frac{2}{3} (x + 1)^{\frac{3}{2}} - x - 1

Add a constant to the solution

= \frac{2}{3} (x + 1)^{\frac{3}{2}} - x - 1 + C

\int \frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} d x

\int ln ∣ x - 1 ∣ d x

f (x) = \frac{x}{1 - x ^{2}}

f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

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