What is the integral of (5x+7)/(x^2+2x-3) ?

The integral of (5x+7)/(x^2+2x-3) is 5/2 ln|x^2+2x-3|-1/2 ln| 1/2 (x+1)+1|+1/2 ln| 1/2 (x+1)-1|+C

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integral of (5x+7)/(x^2+2x-3)

\int \frac{5 x + 7}{x ^{2} + 2 x - 3} d x

\frac{5}{2} 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 + C

Solution steps

\int \frac{5 x + 7}{x ^{2} + 2 x - 3} d x

Complete the square

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

= \int \frac{5 x + 7}{( x + 1 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{5 u + 2}{u ^{2} - 4} d u

= \int \frac{5 u + 2}{u ^{2} - 4} d u

Expand

\frac{5 u + 2}{u ^{2} - 4} : \frac{5 u}{u ^{2} - 4} + \frac{2}{u ^{2} - 4}

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{5 u}{u ^{2} - 4} d u + \int \frac{2}{u ^{2} - 4} d u

\int \frac{5 u}{u ^{2} - 4} d u = \frac{5}{2} ln u^{2} - 4

\int \frac{2}{u ^{2} - 4} d u = - \frac{1}{2} ln \frac{u}{2} + 1 + \frac{1}{2} ln \frac{u}{2} - 1

= \frac{5}{2} ln u^{2} - 4 - \frac{1}{2} ln \frac{u}{2} + 1 + \frac{1}{2} ln \frac{u}{2} - 1

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

= \frac{5}{2} 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 + C

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What is the integral of (5x+7)/(x^2+2x-3) ?
The integral of (5x+7)/(x^2+2x-3) is 5/2 ln|x^2+2x-3|-1/2 ln| 1/2 (x+1)+1|+1/2 ln| 1/2 (x+1)-1|+C