What is the integral of-(3(x-6))/(x^2-17x+72) ?

The integral of-(3(x-6))/(x^2-17x+72) is -6(1/4 ln|4x^2-68x+288|-5/4 (ln|2x-16|-ln|2x-18|))+C

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integral of-(3(x-6))/(x^2-17x+72)

\int - \frac{3 ( x - 6 )}{x ^{2} - 17 x + 72} d x

- 6 (\frac{1}{4} ln 4 x^{2} - 68 x + 288 - \frac{5}{4} (ln ∣ 2 x - 16 ∣ - ln ∣ 2 x - 18 ∣)) + C

Solution steps

\int - \frac{3 ( x - 6 )}{x ^{2} - 17 x + 72} d x

Take the constant out:

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

= - 3 \cdot \int \frac{x - 6}{x ^{2} - 17 x + 72} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 5 u + 6} d u

= - 3 \cdot \int \frac{u}{u ^{2} - 5 u + 6} d u

Complete the square

u^{2} - 5 u + 6 : (u - \frac{5}{2})^{2} - \frac{1}{4}

= - 3 \cdot \int \frac{u}{( u - \frac{5}{2} ) ^{2} - \frac{1}{4}} d u

Apply u-substitution

: \int \frac{2 ( 2 v + 5 )}{4 v ^{2} - 1} d v

= - 3 \cdot \int \frac{2 ( 2 v + 5 )}{4 v ^{2} - 1} d v

Take the constant out:

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

= - 3 \cdot 2 \cdot \int \frac{2 v + 5}{4 v ^{2} - 1} d v

Expand

\frac{2 v + 5}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} + \frac{5}{4 v ^{2} - 1}

Apply the Sum Rule:

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

= - 3 \cdot 2 (\int \frac{2 v}{4 v ^{2} - 1} d v + \int \frac{5}{4 v ^{2} - 1} d v)

\int \frac{2 v}{4 v ^{2} - 1} d v = \frac{1}{4} ln 4 v^{2} - 1

\int \frac{5}{4 v ^{2} - 1} d v = - \frac{5}{4} (ln ∣ 2 v + 1 ∣ - ln ∣ 2 v - 1 ∣)

= - 3 \cdot 2 (\frac{1}{4} ln 4 v^{2} - 1 - \frac{5}{4} (ln ∣ 2 v + 1 ∣ - ln ∣ 2 v - 1 ∣))

Substitute back

= - 3 \cdot 2 (\frac{1}{4} ln 4 (x - 6 - \frac{5}{2})^{2} - 1 - \frac{5}{4} (ln 2 (x - 6 - \frac{5}{2}) + 1 - ln 2 (x - 6 - \frac{5}{2}) - 1))

Simplify

- 3 \cdot 2 (\frac{1}{4} ln 4 (x - 6 - \frac{5}{2})^{2} - 1 - \frac{5}{4} (ln 2 (x - 6 - \frac{5}{2}) + 1 - ln 2 (x - 6 - \frac{5}{2}) - 1)) : - 6 (\frac{1}{4} ln 4 x^{2} - 68 x + 288 - \frac{5}{4} (ln ∣ 2 x - 16 ∣ - ln ∣ 2 x - 18 ∣))

= - 6 (\frac{1}{4} ln 4 x^{2} - 68 x + 288 - \frac{5}{4} (ln ∣ 2 x - 16 ∣ - ln ∣ 2 x - 18 ∣))

Add a constant to the solution

= - 6 (\frac{1}{4} ln 4 x^{2} - 68 x + 288 - \frac{5}{4} (ln ∣ 2 x - 16 ∣ - ln ∣ 2 x - 18 ∣)) + C

\frac{3}{s ( s ^{2} + 6 s )}

x \to 5 lim (- x^{2} + 2 x - 5)

\int_{0}^{10} 400 - 20 t d t

x \to 0 + lim (lo g_{\frac{1}{2}} (x))

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

What is the integral of-(3(x-6))/(x^2-17x+72) ?
The integral of-(3(x-6))/(x^2-17x+72) is -6(1/4 ln|4x^2-68x+288|-5/4 (ln|2x-16|-ln|2x-18|))+C