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

y = (5 - x^{2})^{5}

\int x^{2} + 6 x + 13 d x

f (x) = \frac{1}{x ^{3}}, at x = 1

\frac{\partial}{\partial x} (y^{6} sin (5 x))

arctan (\frac{4}{x})

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