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

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

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integral of x/(1+sqrt(2x+1))

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

\frac{1}{4} (\frac{2}{3} (2 x + 1)^{\frac{3}{2}} - 2 x - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{4 ( 1 + u )} d u

= \int \frac{u - 1}{4 ( 1 + u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{1 + u} d u

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

= \frac{1}{4} \cdot \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

= \frac{1}{4} (\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{1}{4} (\frac{2}{3} u^{\frac{3}{2}} - u)

Substitute back

u = 2 x + 1

= \frac{1}{4} (\frac{2}{3} (2 x + 1)^{\frac{3}{2}} - (2 x + 1))

Simplify

= \frac{1}{4} (\frac{2}{3} (2 x + 1)^{\frac{3}{2}} - 2 x - 1)

Add a constant to the solution

= \frac{1}{4} (\frac{2}{3} (2 x + 1)^{\frac{3}{2}} - 2 x - 1) + C

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