What is the integral of (7x^2+16x-19)/(x^2+2x-3) ?

The integral of (7x^2+16x-19)/(x^2+2x-3) is

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integral of (7x^2+16x-19)/(x^2+2x-3)

\int \frac{7 x ^{2} + 16 x - 19}{x ^{2} + 2 x - 3} d x

7 (x + 1 - \frac{9}{4} ln ∣ x + 3 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + 16 (\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{19}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) + C

Solution steps

\int \frac{7 x ^{2} + 16 x - 19}{x ^{2} + 2 x - 3} d x

Expand

\frac{7 x ^{2} + 16 x - 19}{x ^{2} + 2 x - 3} : \frac{7 x ^{2}}{x ^{2} + 2 x - 3} + \frac{16 x}{x ^{2} + 2 x - 3} - \frac{19}{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{7 x ^{2}}{x ^{2} + 2 x - 3} d x + \int \frac{16 x}{x ^{2} + 2 x - 3} d x - \int \frac{19}{x ^{2} + 2 x - 3} d x

\int \frac{7 x ^{2}}{x ^{2} + 2 x - 3} d x = 7 (x + 1 - \frac{9}{4} ln ∣ x + 3 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

\int \frac{16 x}{x ^{2} + 2 x - 3} d x = 16 (\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{19}{x ^{2} + 2 x - 3} d x = - \frac{19}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1)

= 7 (x + 1 - \frac{9}{4} ln ∣ x + 3 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + 16 (\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{19}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1))

Simplify

7 (x + 1 - \frac{9}{4} ln ∣ x + 3 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + 16 (\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{19}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1)) : 7 (x + 1 - \frac{9}{4} ln ∣ x + 3 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + 16 (\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{19}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)

= 7 (x + 1 - \frac{9}{4} ln ∣ x + 3 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + 16 (\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{19}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)

Add a constant to the solution

= 7 (x + 1 - \frac{9}{4} ln ∣ x + 3 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + 16 (\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{19}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) + C

f (x) = \frac{1}{3 ( x ) ^{2} + 2 ( x )}

f (x) = 2 x^{3} + 42 x^{2} + 288 x + 14

y = 2 - 6 x

\int (3 + x)^{2} d x

f (x) = - x^{2}

What is the integral of (7x^2+16x-19)/(x^2+2x-3) ?
The integral of (7x^2+16x-19)/(x^2+2x-3) is