What is the integral of (-2x^2)/(x^2-2x-48) ?

The integral of (-2x^2)/(x^2-2x-48) is -2(x-1-18/7 ln|x+6|+32/7 ln|x-8|)+C

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integral of (-2x^2)/(x^2-2x-48)

\int \frac{- 2 x ^{2}}{x ^{2} - 2 x - 48} d x

- 2 (x - 1 - \frac{18}{7} ln ∣ x + 6 ∣ + \frac{32}{7} ln ∣ x - 8 ∣) + C

Solution steps

\int \frac{- 2 x ^{2}}{x ^{2} - 2 x - 48} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int \frac{x ^{2}}{x ^{2} - 2 x - 48} d x

Complete the square

x^{2} - 2 x - 48 : (x - 1)^{2} - 49

= - 2 \cdot \int \frac{x ^{2}}{( x - 1 ) ^{2} - 49} d x

Apply u-substitution

= - 2 \cdot \int \frac{( u + 1 ) ^{2}}{u ^{2} - 49} d u

Take the partial fraction of

\frac{( u + 1 ) ^{2}}{u ^{2} - 49} : 1 - \frac{18}{7 ( u + 7 )} + \frac{32}{7 ( u - 7 )}

= - 2 \cdot \int 1 - \frac{18}{7 ( u + 7 )} + \frac{32}{7 ( u - 7 )} 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

= - 2 (\int 1 d u - \int \frac{18}{7 ( u + 7 )} d u + \int \frac{32}{7 ( u - 7 )} d u)

\int 1 d u = u

\int \frac{18}{7 ( u + 7 )} d u = \frac{18}{7} ln ∣ u + 7 ∣

\int \frac{32}{7 ( u - 7 )} d u = \frac{32}{7} ln ∣ u - 7 ∣

= - 2 (u - \frac{18}{7} ln ∣ u + 7 ∣ + \frac{32}{7} ln ∣ u - 7 ∣)

Substitute back

u = x - 1

= - 2 (x - 1 - \frac{18}{7} ln ∣ x - 1 + 7 ∣ + \frac{32}{7} ln ∣ x - 1 - 7 ∣)

Simplify

- 2 (x - 1 - \frac{18}{7} ln ∣ x - 1 + 7 ∣ + \frac{32}{7} ln ∣ x - 1 - 7 ∣) : - 2 (x - 1 - \frac{18}{7} ln ∣ x + 6 ∣ + \frac{32}{7} ln ∣ x - 8 ∣)

= - 2 (x - 1 - \frac{18}{7} ln ∣ x + 6 ∣ + \frac{32}{7} ln ∣ x - 8 ∣)

Add a constant to the solution

= - 2 (x - 1 - \frac{18}{7} ln ∣ x + 6 ∣ + \frac{32}{7} ln ∣ x - 8 ∣) + C

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What is the integral of (-2x^2)/(x^2-2x-48) ?
The integral of (-2x^2)/(x^2-2x-48) is -2(x-1-18/7 ln|x+6|+32/7 ln|x-8|)+C