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

The integral of x/(2(x+1)sqrt(2x+1)) is 1/2 (-2arctan(sqrt(2x+1))+sqrt(2x+1))+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u ( u + 1 )} : \frac{u}{u ( u + 1 )} - \frac{1}{u ( u + 1 )}

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 \frac{1}{2} (\int \frac{u}{u ( u + 1 )} d u - \int \frac{1}{u ( u + 1 )} d u)

\int \frac{u}{u ( u + 1 )} d u = 2 (- arctan (u) + u)

\int \frac{1}{u ( u + 1 )} d u = 2 arctan (u)

= \frac{1}{2} \cdot \frac{1}{2} (2 (- arctan (u) + u) - 2 arctan (u))

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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\frac{d y}{d x} = \frac{( 4 y lo g _{e} ( x ) )}{x}

\frac{1}{( 1 + x ^{2} ) , ( 1 , \frac{1}{2} )}

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What is the integral of x/(2(x+1)sqrt(2x+1)) ?
The integral of x/(2(x+1)sqrt(2x+1)) is 1/2 (-2arctan(sqrt(2x+1))+sqrt(2x+1))+C