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

The integral of (7(4-x^2))/((x-4)(x-2)(x+3)) is -6ln|x-4|-ln|x+3|+C

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

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

- 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣ + C

Solution steps

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

Take the constant out:

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

= 7 \cdot \int \frac{4 - x ^{2}}{( x - 4 ) ( x - 2 ) ( x + 3 )} d x

Apply u-substitution

: \int - \frac{u - 1}{u ( u - 7 )} d u

= 7 \cdot \int - \frac{u - 1}{u ( u - 7 )} d u

Take the constant out:

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

= 7 (- \int \frac{u - 1}{u ( u - 7 )} d u)

Expand

\frac{u - 1}{u ( u - 7 )} : \frac{u}{u ( u - 7 )} - \frac{1}{u ( u - 7 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 (- (\int \frac{u}{u ( u - 7 )} d u - \int \frac{1}{u ( u - 7 )} d u))

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

\int \frac{1}{u ( u - 7 )} d u = - \frac{1}{7} ln ∣ u ∣ + \frac{1}{7} ln ∣ u - 7 ∣

= 7 (- (ln ∣ u - 7 ∣ - (- \frac{1}{7} ln ∣ u ∣ + \frac{1}{7} ln ∣ u - 7 ∣)))

Substitute back

u = x + 3

= 7 (- (ln ∣ x + 3 - 7 ∣ - (- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x + 3 - 7 ∣)))

Simplify

7 (- (ln ∣ x + 3 - 7 ∣ - (- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x + 3 - 7 ∣))) : - 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣

= - 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣

Add a constant to the solution

= - 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣ + C

\int \frac{1}{5} sin (5 x) d x

\frac{d}{d t} (- 4 t^{3})

\int - \frac{1}{8} sin (2 x) d x

x \to 1 lim (lo g_{4} (x))

x \to 8 + lim (\frac{1}{x - 8} - \frac{1}{x - 8})

What is the integral of (7(4-x^2))/((x-4)(x-2)(x+3)) ?
The integral of (7(4-x^2))/((x-4)(x-2)(x+3)) is -6ln|x-4|-ln|x+3|+C