What is the integral of sqrt(1+1/(4x)) ?

The integral of sqrt(1+1/(4x)) is 1/4 (2sqrt(x)sqrt(4x+1)+ln|sqrt(4x+1)+2sqrt(x)|)+C

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integral of sqrt(1+1/(4x))

\int 1 + \frac{1}{4 x} d x

\frac{1}{4} (2 x 4 x + 1 + ln 4 x + 1 + 2 x) + C

Solution steps

\int 1 + \frac{1}{4 x} d x

Apply u-substitution

: \int \frac{u + 1}{4 u} d u

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

Substitute back

u = 4 x

= \frac{1}{4} (\frac{4 x + 1}{4 x} \cdot 4 x - (- ln 4 x + 1 + 4 x))

Simplify

\frac{1}{4} (\frac{4 x + 1}{4 x} \cdot 4 x - (- ln 4 x + 1 + 4 x)) : \frac{1}{4} (2 x 4 x + 1 + ln 4 x + 1 + 2 x)

= \frac{1}{4} (2 x 4 x + 1 + ln 4 x + 1 + 2 x)

Add a constant to the solution

= \frac{1}{4} (2 x 4 x + 1 + ln 4 x + 1 + 2 x) + C

\frac{d}{d x} (cot (3 x^{4} - 2 x^{3} - 3))

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\frac{\partial}{\partial y} (2 y e^{2 z})

What is the integral of sqrt(1+1/(4x)) ?
The integral of sqrt(1+1/(4x)) is 1/4 (2sqrt(x)sqrt(4x+1)+ln|sqrt(4x+1)+2sqrt(x)|)+C