What is the integral of (x^2)/(x^2+2x-15) ?

The integral of (x^2)/(x^2+2x-15) is x+1-25/8 ln|x+5|+9/8 ln|x-3|+C

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integral of (x^2)/(x^2+2x-15)

\int \frac{x ^{2}}{x ^{2} + 2 x - 15} d x

x + 1 - \frac{25}{8} ln ∣ x + 5 ∣ + \frac{9}{8} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2}}{x ^{2} + 2 x - 15} d x

Complete the square

x^{2} + 2 x - 15 : (x + 1)^{2} - 16

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

Apply u-substitution

: \int \frac{( u - 1 ) ^{2}}{u ^{2} - 16} d u

= \int \frac{( u - 1 ) ^{2}}{u ^{2} - 16} d u

Take the partial fraction of

\frac{( u - 1 ) ^{2}}{u ^{2} - 16} : 1 - \frac{25}{8 ( u + 4 )} + \frac{9}{8 ( u - 4 )}

= \int 1 - \frac{25}{8 ( u + 4 )} + \frac{9}{8 ( u - 4 )} 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 1 d u - \int \frac{25}{8 ( u + 4 )} d u + \int \frac{9}{8 ( u - 4 )} d u

\int 1 d u = u

\int \frac{25}{8 ( u + 4 )} d u = \frac{25}{8} ln ∣ u + 4 ∣

\int \frac{9}{8 ( u - 4 )} d u = \frac{9}{8} ln ∣ u - 4 ∣

= u - \frac{25}{8} ln ∣ u + 4 ∣ + \frac{9}{8} ln ∣ u - 4 ∣

Substitute back

u = x + 1

= x + 1 - \frac{25}{8} ln ∣ x + 1 + 4 ∣ + \frac{9}{8} ln ∣ x + 1 - 4 ∣

Simplify

x + 1 - \frac{25}{8} ln ∣ x + 1 + 4 ∣ + \frac{9}{8} ln ∣ x + 1 - 4 ∣ : x + 1 - \frac{25}{8} ln ∣ x + 5 ∣ + \frac{9}{8} ln ∣ x - 3 ∣

= x + 1 - \frac{25}{8} ln ∣ x + 5 ∣ + \frac{9}{8} ln ∣ x - 3 ∣

Add a constant to the solution

= x + 1 - \frac{25}{8} ln ∣ x + 5 ∣ + \frac{9}{8} ln ∣ x - 3 ∣ + C

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What is the integral of (x^2)/(x^2+2x-15) ?
The integral of (x^2)/(x^2+2x-15) is x+1-25/8 ln|x+5|+9/8 ln|x-3|+C