What is the integral of (-3x-3)/(sqrt(x^2+2x-4)) ?

The integral of (-3x-3)/(sqrt(x^2+2x-4)) is -3sqrt(x^2+2x-4)+C

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integral of (-3x-3)/(sqrt(x^2+2x-4))

\int \frac{- 3 x - 3}{x ^{2} + 2 x - 4} d x

- 3 x^{2} + 2 x - 4 + C

Solution steps

\int \frac{- 3 x - 3}{x ^{2} + 2 x - 4} d x

Expand

\frac{- 3 x - 3}{x ^{2} + 2 x - 4} : - \frac{3 x}{x ^{2} + 2 x - 4} - \frac{3}{x ^{2} + 2 x - 4}

= \int - \frac{3 x}{x ^{2} + 2 x - 4} - \frac{3}{x ^{2} + 2 x - 4} d x

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 x}{x ^{2} + 2 x - 4} d x - \int \frac{3}{x ^{2} + 2 x - 4} d x

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

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

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

Simplify

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

= - 3 x^{2} + 2 x - 4

Add a constant to the solution

= - 3 x^{2} + 2 x - 4 + C

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What is the integral of (-3x-3)/(sqrt(x^2+2x-4)) ?
The integral of (-3x-3)/(sqrt(x^2+2x-4)) is -3sqrt(x^2+2x-4)+C