What is the integral of (4x+7)/(x^2+2x-3) ?

The integral of (4x+7)/(x^2+2x-3) is 5/4 ln|x+3|+11/4 ln|x-1|+C

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integral of (4x+7)/(x^2+2x-3)

\int \frac{4 x + 7}{x ^{2} + 2 x - 3} d x

\frac{5}{4} ln ∣ x + 3 ∣ + \frac{11}{4} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{4 x + 7}{x ^{2} + 2 x - 3} d x

Complete the square

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

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{4 u + 3}{u ^{2} - 4} : \frac{5}{4 ( u + 2 )} + \frac{11}{4 ( u - 2 )}

= \int \frac{5}{4 ( u + 2 )} + \frac{11}{4 ( u - 2 )} d u

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}{4 ( u + 2 )} d u + \int \frac{11}{4 ( u - 2 )} d u

\int \frac{5}{4 ( u + 2 )} d u = \frac{5}{4} ln ∣ u + 2 ∣

\int \frac{11}{4 ( u - 2 )} d u = \frac{11}{4} ln ∣ u - 2 ∣

= \frac{5}{4} ln ∣ u + 2 ∣ + \frac{11}{4} ln ∣ u - 2 ∣

Substitute back

u = x + 1

= \frac{5}{4} ln ∣ x + 1 + 2 ∣ + \frac{11}{4} ln ∣ x + 1 - 2 ∣

Simplify

\frac{5}{4} ln ∣ x + 1 + 2 ∣ + \frac{11}{4} ln ∣ x + 1 - 2 ∣ : \frac{5}{4} ln ∣ x + 3 ∣ + \frac{11}{4} ln ∣ x - 1 ∣

= \frac{5}{4} ln ∣ x + 3 ∣ + \frac{11}{4} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{5}{4} ln ∣ x + 3 ∣ + \frac{11}{4} ln ∣ x - 1 ∣ + C

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What is the integral of (4x+7)/(x^2+2x-3) ?
The integral of (4x+7)/(x^2+2x-3) is 5/4 ln|x+3|+11/4 ln|x-1|+C