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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} \cdot 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

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

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

\int 1 d u = u

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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