What is the integral of (8-3x)/((x+1)(x^2-4x+6)) ?

The integral of (8-3x)/((x+1)(x^2-4x+6)) is -1/2 ln|9x^2-36x+54|+ln|3x+3|+C

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integral of (8-3x)/((x+1)(x^2-4x+6))

\int \frac{8 - 3 x}{( x + 1 ) ( x ^{2} - 4 x + 6 )} d x

- \frac{1}{2} ln 9 x^{2} - 36 x + 54 + ln ∣ 3 x + 3 ∣ + C

Solution steps

\int \frac{8 - 3 x}{( x + 1 ) ( x ^{2} - 4 x + 6 )} d x

Apply u-substitution

: \int - \frac{9 u}{( u ^{2} - 4 u + 22 ) ( - u + 11 )} d u

= \int - \frac{9 u}{( u ^{2} - 4 u + 22 ) ( - u + 11 )} d u

Take the constant out:

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

= - 9 \cdot \int \frac{u}{( u ^{2} - 4 u + 22 ) ( - u + 11 )} d u

Take the partial fraction of

\frac{u}{( u ^{2} - 4 u + 22 ) ( - u + 11 )} : \frac{u - 2}{9 ( u ^{2} - 4 u + 22 )} - \frac{1}{9 ( u - 11 )}

= - 9 \cdot \int \frac{u - 2}{9 ( u ^{2} - 4 u + 22 )} - \frac{1}{9 ( u - 11 )} 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

= - 9 (\int \frac{u - 2}{9 ( u ^{2} - 4 u + 22 )} d u - \int \frac{1}{9 ( u - 11 )} d u)

\int \frac{u - 2}{9 ( u ^{2} - 4 u + 22 )} d u = \frac{1}{18} ln u^{2} - 4 u + 22

\int \frac{1}{9 ( u - 11 )} d u = \frac{1}{9} ln ∣ u - 11 ∣

= - 9 (\frac{1}{18} ln u^{2} - 4 u + 22 - \frac{1}{9} ln ∣ u - 11 ∣)

Substitute back

u = 8 - 3 x

= - 9 (\frac{1}{18} ln (8 - 3 x)^{2} - 4 (8 - 3 x) + 22 - \frac{1}{9} ln ∣ 8 - 3 x - 11 ∣)

Simplify

- 9 (\frac{1}{18} ln (8 - 3 x)^{2} - 4 (8 - 3 x) + 22 - \frac{1}{9} ln ∣ 8 - 3 x - 11 ∣) : - \frac{1}{2} ln 9 x^{2} - 36 x + 54 + ln ∣ 3 x + 3 ∣

= - \frac{1}{2} ln 9 x^{2} - 36 x + 54 + ln ∣ 3 x + 3 ∣

Add a constant to the solution

= - \frac{1}{2} ln 9 x^{2} - 36 x + 54 + ln ∣ 3 x + 3 ∣ + C

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What is the integral of (8-3x)/((x+1)(x^2-4x+6)) ?
The integral of (8-3x)/((x+1)(x^2-4x+6)) is -1/2 ln|9x^2-36x+54|+ln|3x+3|+C