What is the integral of (4x+5)/(x^2-11x+30) ?

The integral of (4x+5)/(x^2-11x+30) is 2ln|4x^2-44x+120|-27ln|2x-10|+27ln|2x-12|+C

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integral of (4x+5)/(x^2-11x+30)

\int \frac{4 x + 5}{x ^{2} - 11 x + 30} d x

2 ln 4 x^{2} - 44 x + 120 - 27 ln ∣ 2 x - 10 ∣ + 27 ln ∣ 2 x - 12 ∣ + C

Solution steps

\int \frac{4 x + 5}{x ^{2} - 11 x + 30} d x

Complete the square

x^{2} - 11 x + 30 : (x - \frac{11}{2})^{2} - \frac{1}{4}

= \int \frac{4 x + 5}{( x - \frac{11}{2} ) ^{2} - \frac{1}{4}} d x

Apply u-substitution

= \int \frac{4 ( 4 u + 27 )}{4 u ^{2} - 1} d u

Take the constant out:

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

= 4 \cdot \int \frac{4 u + 27}{4 u ^{2} - 1} d u

Expand

\frac{4 u + 27}{4 u ^{2} - 1} : \frac{4 u}{4 u ^{2} - 1} + \frac{27}{4 u ^{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

= 4 (\int \frac{4 u}{4 u ^{2} - 1} d u + \int \frac{27}{4 u ^{2} - 1} d u)

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

\int \frac{27}{4 u ^{2} - 1} d u = - \frac{27}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 4 (\frac{1}{2} ln 4 u^{2} - 1 - \frac{27}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣))

Substitute back

u = x - \frac{11}{2}

= 4 (\frac{1}{2} ln 4 (x - \frac{11}{2})^{2} - 1 - \frac{27}{4} (ln 2 (x - \frac{11}{2}) + 1 - ln 2 (x - \frac{11}{2}) - 1))

Simplify

4 (\frac{1}{2} ln 4 (x - \frac{11}{2})^{2} - 1 - \frac{27}{4} (ln 2 (x - \frac{11}{2}) + 1 - ln 2 (x - \frac{11}{2}) - 1)) : 2 ln 4 x^{2} - 44 x + 120 - 27 ln ∣ 2 x - 10 ∣ + 27 ln ∣ 2 x - 12 ∣

= 2 ln 4 x^{2} - 44 x + 120 - 27 ln ∣ 2 x - 10 ∣ + 27 ln ∣ 2 x - 12 ∣

Add a constant to the solution

= 2 ln 4 x^{2} - 44 x + 120 - 27 ln ∣ 2 x - 10 ∣ + 27 ln ∣ 2 x - 12 ∣ + C

What is the integral of (4x+5)/(x^2-11x+30) ?
The integral of (4x+5)/(x^2-11x+30) is 2ln|4x^2-44x+120|-27ln|2x-10|+27ln|2x-12|+C