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

The integral of sqrt(1+1/(2x)) is 1/2 (sqrt((2(2x+1))/x)x+ln|sqrt(2x+1)+sqrt(2)sqrt(x)|)+C

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

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

\frac{1}{2} (\frac{2 ( 2 x + 1 )}{x} x + ln 2 x + 1 + 2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 1}{2 u} d u

= \int \frac{u + 1}{2 u} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

\int - \frac{1}{2 u ^{\frac{1}{2}} u + 1} d u = - ln u + 1 + u

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

Substitute back

u = 2 x

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

Simplify

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

= \frac{1}{2} (\frac{2 ( 2 x + 1 )}{x} x + ln 2 x + 1 + 2 x)

Add a constant to the solution

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

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