What is the integral of (2x-1)/(x^2-8x+15) ?

The integral of (2x-1)/(x^2-8x+15) is ln|x^2-8x+15|-7(1/2 ln|x-3|-1/2 ln|x-5|)+C

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integral of (2x-1)/(x^2-8x+15)

\int \frac{2 x - 1}{x ^{2} - 8 x + 15} d x

ln x^{2} - 8 x + 15 - 7 (\frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 5 ∣) + C

Solution steps

\int \frac{2 x - 1}{x ^{2} - 8 x + 15} d x

Complete the square

x^{2} - 8 x + 15 : (x - 4)^{2} - 1

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

Apply u-substitution

: \int \frac{2 u + 7}{u ^{2} - 1} d u

= \int \frac{2 u + 7}{u ^{2} - 1} d u

Expand

\frac{2 u + 7}{u ^{2} - 1} : \frac{2 u}{u ^{2} - 1} + \frac{7}{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

= \int \frac{2 u}{u ^{2} - 1} d u + \int \frac{7}{u ^{2} - 1} d u

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

\int \frac{7}{u ^{2} - 1} d u = - 7 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

= ln u^{2} - 1 - 7 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

Substitute back

u = x - 4

= ln (x - 4)^{2} - 1 - 7 (\frac{1}{2} ln ∣ x - 4 + 1 ∣ - \frac{1}{2} ln ∣ x - 4 - 1 ∣)

Simplify

ln (x - 4)^{2} - 1 - 7 (\frac{1}{2} ln ∣ x - 4 + 1 ∣ - \frac{1}{2} ln ∣ x - 4 - 1 ∣) : ln x^{2} - 8 x + 15 - 7 (\frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 5 ∣)

= ln x^{2} - 8 x + 15 - 7 (\frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 5 ∣)

Add a constant to the solution

= ln x^{2} - 8 x + 15 - 7 (\frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 5 ∣) + C

\int \frac{x + 4}{5 x + 5} d x

\int sin^{5} (2 t) cos^{2} (2 t) d t

\frac{d}{d x} (\frac{1}{8 x})

\int - t e^{t} d t

\int \frac{1}{10 - x ^{2}} d x

What is the integral of (2x-1)/(x^2-8x+15) ?
The integral of (2x-1)/(x^2-8x+15) is ln|x^2-8x+15|-7(1/2 ln|x-3|-1/2 ln|x-5|)+C