What is the integral of (6x+6)/(x^2+2x-3) ?

The integral of (6x+6)/(x^2+2x-3) is

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integral of (6x+6)/(x^2+2x-3)

\int \frac{6 x + 6}{x ^{2} + 2 x - 3} d x

6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - 3 (\frac{1}{2} ln \frac{1}{2} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{2} (x + 1) - 1) + C

Solution steps

\int \frac{6 x + 6}{x ^{2} + 2 x - 3} d x

Expand

\frac{6 x + 6}{x ^{2} + 2 x - 3} : \frac{6 x}{x ^{2} + 2 x - 3} + \frac{6}{x ^{2} + 2 x - 3}

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{6 x}{x ^{2} + 2 x - 3} d x + \int \frac{6}{x ^{2} + 2 x - 3} d x

\int \frac{6 x}{x ^{2} + 2 x - 3} d x = 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1))

\int \frac{6}{x ^{2} + 2 x - 3} d x = - 3 (\frac{1}{2} ln \frac{x + 1}{2} + 1 - \frac{1}{2} ln \frac{x + 1}{2} - 1)

= 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - 3 (\frac{1}{2} ln \frac{x + 1}{2} + 1 - \frac{1}{2} ln \frac{x + 1}{2} - 1)

Simplify

6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - 3 (\frac{1}{2} ln \frac{x + 1}{2} + 1 - \frac{1}{2} ln \frac{x + 1}{2} - 1) : 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - 3 (\frac{1}{2} ln \frac{1}{2} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{2} (x + 1) - 1)

= 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - 3 (\frac{1}{2} ln \frac{1}{2} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{2} (x + 1) - 1)

Add a constant to the solution

= 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - 3 (\frac{1}{2} ln \frac{1}{2} (x + 1) + 1 - \frac{1}{2} ln \frac{1}{2} (x + 1) - 1) + C

What is the integral of (6x+6)/(x^2+2x-3) ?
The integral of (6x+6)/(x^2+2x-3) is