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

The integral of (2x-3)/(x^2-8x+7) is 1/6 ln|x-1|+11/6 ln|x-7|+C

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

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

\frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x - 7 ∣ + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{2 u + 5}{u ^{2} - 9} : \frac{1}{6 ( u + 3 )} + \frac{11}{6 ( u - 3 )}

= \int \frac{1}{6 ( u + 3 )} + \frac{11}{6 ( u - 3 )} 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

= \int \frac{1}{6 ( u + 3 )} d u + \int \frac{11}{6 ( u - 3 )} d u

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

\int \frac{11}{6 ( u - 3 )} d u = \frac{11}{6} ln ∣ u - 3 ∣

= \frac{1}{6} ln ∣ u + 3 ∣ + \frac{11}{6} ln ∣ u - 3 ∣

Substitute back

u = x - 4

= \frac{1}{6} ln ∣ x - 4 + 3 ∣ + \frac{11}{6} ln ∣ x - 4 - 3 ∣

Simplify

\frac{1}{6} ln ∣ x - 4 + 3 ∣ + \frac{11}{6} ln ∣ x - 4 - 3 ∣ : \frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x - 7 ∣

= \frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x - 7 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x - 1 ∣ + \frac{11}{6} ln ∣ x - 7 ∣ + C

\int t t - 6 d t

\int 9 x (6 - x) d x

\int \frac{x}{( x ^{2} + 9 )} d x

\int 2 u d u

\int t^{2} + t^{4} d t

What is the integral of (2x-3)/(x^2-8x+7) ?
The integral of (2x-3)/(x^2-8x+7) is 1/6 ln|x-1|+11/6 ln|x-7|+C