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

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

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integral of ((3+x))/(1-x^2)

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

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

Solution steps

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

Expand

\frac{3 + x}{1 - x ^{2}} : \frac{3}{1 - x ^{2}} + \frac{x}{1 - x ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Add a constant to the solution

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

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