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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int e^{- 2 k x} d x

y = 7 + 9 x

\frac{d}{d x} (5 sec (x) - 10 cos (x))

(x^{2} + 3) y^{^{'}} = x y

(x^{3} - 9 x)^{12}

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