What is the integral of (5x-11)/(x^2+10x+9) ?

The integral of (5x-11)/(x^2+10x+9) is 5/2 ln(|x+9||x+1|)-36(-1/8 ln|x+9|+1/8 ln|x+1|)+C

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integral of (5x-11)/(x^2+10x+9)

\int \frac{5 x - 11}{x ^{2} + 10 x + 9} d x

\frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣) + C

Solution steps

\int \frac{5 x - 11}{x ^{2} + 10 x + 9} d x

Complete the square

x^{2} + 10 x + 9 : (x + 5)^{2} - 16

= \int \frac{5 x - 11}{( x + 5 ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{5 u - 36}{u ^{2} - 16} d u

= \int \frac{5 u - 36}{u ^{2} - 16} d u

Factor

\frac{5 u - 36}{u ^{2} - 16}

= \int \frac{5 u - 36}{( u + 4 ) ( u - 4 )} d u

Expand

\frac{5 u - 36}{( u + 4 ) ( u - 4 )} : \frac{5 u}{( u + 4 ) ( u - 4 )} - \frac{36}{( u + 4 ) ( u - 4 )}

Apply the Sum Rule:

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

= \int \frac{5 u}{( u + 4 ) ( u - 4 )} d u - \int \frac{36}{( u + 4 ) ( u - 4 )} d u

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

\int \frac{36}{( u + 4 ) ( u - 4 )} d u = 36 (- \frac{1}{8} ln ∣ u + 4 ∣ + \frac{1}{8} ln ∣ u - 4 ∣)

= \frac{5}{2} ln (∣ u + 4 ∣ ∣ u - 4 ∣) - 36 (- \frac{1}{8} ln ∣ u + 4 ∣ + \frac{1}{8} ln ∣ u - 4 ∣)

Substitute back

u = x + 5

= \frac{5}{2} ln (∣ x + 5 + 4 ∣ ∣ x + 5 - 4 ∣) - 36 (- \frac{1}{8} ln ∣ x + 5 + 4 ∣ + \frac{1}{8} ln ∣ x + 5 - 4 ∣)

Simplify

\frac{5}{2} ln (∣ x + 5 + 4 ∣ ∣ x + 5 - 4 ∣) - 36 (- \frac{1}{8} ln ∣ x + 5 + 4 ∣ + \frac{1}{8} ln ∣ x + 5 - 4 ∣) : \frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣)

= \frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣)

Add a constant to the solution

= \frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣) + C

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What is the integral of (5x-11)/(x^2+10x+9) ?
The integral of (5x-11)/(x^2+10x+9) is 5/2 ln(|x+9||x+1|)-36(-1/8 ln|x+9|+1/8 ln|x+1|)+C