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

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

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

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

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

Solution steps

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

Simplify

\frac{4 x + 1}{4 x} : \frac{4 x + 1}{2 x}

= \int \frac{4 x + 1}{2 x} 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{4 x + 1}{x} d x

Apply Integration By Parts

: 2 x 4 x + 1 - \int \frac{4 x}{4 x + 1} d x

= \frac{1}{2} (2 x 4 x + 1 - \int \frac{4 x}{4 x + 1} d x)

\int \frac{4 x}{4 x + 1} d x = - \frac{1}{2} ln 2 x + 4 x + 1 + x 4 x + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1 - 3 x}{1 + 3 x})

n = 1 \sum \infty \frac{1}{n ^{2}} - \frac{1}{n}

n = 2 \sum \infty \frac{n}{n + 1}

x \to 5 lim (x^{2} - 25)

\int \frac{2}{( x ^{2} - 3 x + 2 )} d x

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