What is the integral of 1/(4x-x^2) ?

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

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integral of 1/(4x-x^2)

\int \frac{1}{4 x - x ^{2}} d x

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

Solution steps

\int \frac{1}{4 x - x ^{2}} d x

Complete the square

4 x - x^{2} : - (x - 2)^{2} + 4

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

Apply u-substitution

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

= \int \frac{1}{- u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (arctan (2 x + 1))

x \to \infty lim (e^{1})

\frac{d}{d x} (x^{2} - 4 x + 9)

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x} + x^{2}

\frac{d y}{d x} = 13 x^{12} y

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