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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int 2 (u^{2} - 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int u^{2} - 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

= 2 (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (y - 1)

x \to - \infty lim (5)

\int (e^{- x})^{2} d x

h (x) = - x^{3} e^{x}

in v erse l a pl a ce \frac{s ^{2} + 2 s + 3}{( s + 1 ) ^{3}}

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