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

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

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integral of 2/(2-x^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{x}{2}

= 2 \cdot \frac{1}{2} \frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2}

Simplify

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

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

Add a constant to the solution

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

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