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

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

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

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

\frac{1}{2} ln ∣ 4 x - 1 ∣ + \frac{1}{2} ln 2 x + 1 - \frac{1}{2} ln 2 x - 1 + C

Solution steps

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

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

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

Expand

\frac{2 x - x}{( 2 x ) ^{2} - ( x ) ^{2}} : \frac{2 x}{( 2 x ) ^{2} - ( x ) ^{2}} - \frac{x}{( 2 x ) ^{2} - ( x ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{2 x}{( 2 x ) ^{2} - ( x ) ^{2}} d x = \frac{1}{2} ln ∣ 4 x - 1 ∣

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

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} ln ∣ 4 x - 1 ∣ + \frac{1}{2} ln 2 x + 1 - \frac{1}{2} ln 2 x - 1 + C

\int r e^{\frac{r}{3}} d r

\frac{\partial}{\partial x} (x^{2} + y^{2} - y^{2} x - 2)

x \to \infty + lim (x^{3})

x \to 0 - lim (18 - \frac{4}{x ^{3}})

\frac{d}{d x} (e^{x s i n h (x)})

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