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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= 2 (- \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

= 2 (- \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}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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