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

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

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

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

\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) + \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{x - 2}{x ^{2} + 2 x - 3} d x

Expand

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

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

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

Simplify

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

= \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) + \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

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

\frac{d y}{d x} = 3 x y

\int_{0}^{1} 14 cos (x^{2}) d x

\frac{2}{( s + 2 ) ( s + 3 )}

\int_{0}^{2 π} x d x

t + t