What is the integral of ((x+2))/(x^2-6x+5) ?

The integral of ((x+2))/(x^2-6x+5) is 1/2 ln|x^2-6x+5|-5/4 (ln| 1/2 (x-3)+1|-ln| 1/2 (x-3)-1|)+C

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integral of ((x+2))/(x^2-6x+5)

\int \frac{( x + 2 )}{x ^{2} - 6 x + 5} d x

\frac{1}{2} ln x^{2} - 6 x + 5 - \frac{5}{4} (ln \frac{1}{2} (x - 3) + 1 - ln \frac{1}{2} (x - 3) - 1) + C

Solution steps

\int \frac{x + 2}{x ^{2} - 6 x + 5} d x

Complete the square

x^{2} - 6 x + 5 : (x - 3)^{2} - 4

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

Apply u-substitution

: \int \frac{u + 5}{u ^{2} - 4} d u

= \int \frac{u + 5}{u ^{2} - 4} d u

Expand

\frac{u + 5}{u ^{2} - 4} : \frac{u}{u ^{2} - 4} + \frac{5}{u ^{2} - 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{u}{u ^{2} - 4} d u + \int \frac{5}{u ^{2} - 4} d u

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

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

= \frac{1}{2} ln u^{2} - 4 - \frac{5}{4} (ln \frac{u}{2} + 1 - ln \frac{u}{2} - 1)

Substitute back

u = x - 3

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

Simplify

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

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

Add a constant to the solution

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

sin (1)

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

\int \frac{x + 4}{x ^{2} + 4 x + 4} d x

\int \frac{3 sin ( x )}{cos ^{2} ( x )} d x

\frac{\partial}{\partial y} (ln (x + y))

What is the integral of ((x+2))/(x^2-6x+5) ?
The integral of ((x+2))/(x^2-6x+5) is 1/2 ln|x^2-6x+5|-5/4 (ln| 1/2 (x-3)+1|-ln| 1/2 (x-3)-1|)+C