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

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

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

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

- ln 1 - x + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{1 - u} d u

= \int \frac{1}{1 - u} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

= - ln 1 - x

Add a constant to the solution

= - ln 1 - x + C

x \to 0 lim (ln (e^{x} - 1))

\frac{1}{( x - 1 )}, at x = 2

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

\frac{\partial}{\partial v} ((u^{2} v - v^{3})^{5})

(- 3, 0), (0, - 3)

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