What is the integral of 3/(1-x^2) ?

The integral of 3/(1-x^2) is 3((ln|x+1|)/2-(ln|x-1|)/2)+C

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integral of 3/(1-x^2)

\int \frac{3}{1 - x ^{2}} d x

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

Solution steps

\int \frac{3}{1 - x ^{2}} d x

Take the constant out:

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

= 3 \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}

= 3 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})

Add a constant to the solution

= 3 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}) + C

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x^{2} y^{2})

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

x^{2} + x^{3}

\int e^{t - s} d s

What is the integral of 3/(1-x^2) ?
The integral of 3/(1-x^2) is 3((ln|x+1|)/2-(ln|x-1|)/2)+C