What is the integral of (5x-3)/(x^2+8x+7) ?

The integral of (5x-3)/(x^2+8x+7) is 5/2 ln(|x+7||x+1|)-23(-1/6 ln|x+7|+1/6 ln|x+1|)+C

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integral of (5x-3)/(x^2+8x+7)

\int \frac{5 x - 3}{x ^{2} + 8 x + 7} d x

\frac{5}{2} ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 23 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

Solution steps

\int \frac{5 x - 3}{x ^{2} + 8 x + 7} d x

Complete the square

x^{2} + 8 x + 7 : (x + 4)^{2} - 9

= \int \frac{5 x - 3}{( x + 4 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{5 u - 23}{u ^{2} - 9} d u

= \int \frac{5 u - 23}{u ^{2} - 9} d u

Factor

\frac{5 u - 23}{u ^{2} - 9}

= \int \frac{5 u - 23}{( u + 3 ) ( u - 3 )} d u

Expand

\frac{5 u - 23}{( u + 3 ) ( u - 3 )} : \frac{5 u}{( u + 3 ) ( u - 3 )} - \frac{23}{( u + 3 ) ( u - 3 )}

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 + 3 ) ( u - 3 )} d u - \int \frac{23}{( u + 3 ) ( u - 3 )} d u

\int \frac{5 u}{( u + 3 ) ( u - 3 )} d u = \frac{5}{2} ln (∣ u + 3 ∣ ∣ u - 3 ∣)

\int \frac{23}{( u + 3 ) ( u - 3 )} d u = 23 (- \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣)

= \frac{5}{2} ln (∣ u + 3 ∣ ∣ u - 3 ∣) - 23 (- \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣)

Substitute back

u = x + 4

= \frac{5}{2} ln (∣ x + 4 + 3 ∣ ∣ x + 4 - 3 ∣) - 23 (- \frac{1}{6} ln ∣ x + 4 + 3 ∣ + \frac{1}{6} ln ∣ x + 4 - 3 ∣)

Simplify

\frac{5}{2} ln (∣ x + 4 + 3 ∣ ∣ x + 4 - 3 ∣) - 23 (- \frac{1}{6} ln ∣ x + 4 + 3 ∣ + \frac{1}{6} ln ∣ x + 4 - 3 ∣) : \frac{5}{2} ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 23 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

= \frac{5}{2} ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 23 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

Add a constant to the solution

= \frac{5}{2} ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 23 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

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What is the integral of (5x-3)/(x^2+8x+7) ?
The integral of (5x-3)/(x^2+8x+7) is 5/2 ln(|x+7||x+1|)-23(-1/6 ln|x+7|+1/6 ln|x+1|)+C