What is the integral of (x-5)/(x^2+2x-3) ?

The integral of (x-5)/(x^2+2x-3) is

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integral of (x-5)/(x^2+2x-3)

\int \frac{x - 5}{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{5}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) + C

Solution steps

\int \frac{x - 5}{x ^{2} + 2 x - 3} d x

Expand

\frac{x - 5}{x ^{2} + 2 x - 3} : \frac{x}{x ^{2} + 2 x - 3} - \frac{5}{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{5}{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{5}{x ^{2} + 2 x - 3} d x = - \frac{5}{4} (ln \frac{x + 1}{2} + 1 - 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{5}{4} (ln \frac{x + 1}{2} + 1 - 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{5}{4} (ln \frac{x + 1}{2} + 1 - 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{5}{4} (ln \frac{1}{2} (x + 1) + 1 - 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{5}{4} (ln \frac{1}{2} (x + 1) + 1 - 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{5}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) + C

f (x) = \frac{3}{5 x + 2}, at x = - 1

y = x, y = x^{3}

\int (2 x + 1)^{\frac{5}{6}} d x

\frac{\partial}{\partial x} (32 x + 13 y + 90)

\frac{d}{d t} (ln (a + v (t) t))