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

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

Popular >

integral of 2/(1-x^2)

integral

\frac{2}{1 - x ^{2}}

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Use the common integral:

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

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

Simplify

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

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

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

\frac{ln ( x )}{( 1 + x ^{2} )}

\frac{( x + 5 ) ^{3}}{2}

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

\frac{10 t}{( t + 1 ) ^{2}}

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