What is the integral of (x+2)/(x^2-6x+34) ?

The integral of (x+2)/(x^2-6x+34) is 1/2 ln|x^2-6x+34|+arctan(1/5 (x-3))+C

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integral of (x+2)/(x^2-6x+34)

\int \frac{x + 2}{x ^{2} - 6 x + 34} d x

\frac{1}{2} ln x^{2} - 6 x + 34 + arctan (\frac{1}{5} (x - 3)) + C

Solution steps

\int \frac{x + 2}{x ^{2} - 6 x + 34} d x

Complete the square

x^{2} - 6 x + 34 : (x - 3)^{2} + 25

= \int \frac{x + 2}{( x - 3 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{u + 5}{u ^{2} + 25} d u

= \int \frac{u + 5}{u ^{2} + 25} d u

Expand

\frac{u + 5}{u ^{2} + 25} : \frac{u}{u ^{2} + 25} + \frac{5}{u ^{2} + 25}

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} + 25} d u + \int \frac{5}{u ^{2} + 25} d u

\int \frac{u}{u ^{2} + 25} d u = \frac{1}{2} ln u^{2} + 25

\int \frac{5}{u ^{2} + 25} d u = arctan (\frac{u}{5})

= \frac{1}{2} ln u^{2} + 25 + arctan (\frac{u}{5})

Substitute back

u = x - 3

= \frac{1}{2} ln (x - 3)^{2} + 25 + arctan (\frac{x - 3}{5})

Simplify

\frac{1}{2} ln (x - 3)^{2} + 25 + arctan (\frac{x - 3}{5}) : \frac{1}{2} ln x^{2} - 6 x + 34 + arctan (\frac{1}{5} (x - 3))

= \frac{1}{2} ln x^{2} - 6 x + 34 + arctan (\frac{1}{5} (x - 3))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 6 x + 34 + arctan (\frac{1}{5} (x - 3)) + C

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What is the integral of (x+2)/(x^2-6x+34) ?
The integral of (x+2)/(x^2-6x+34) is 1/2 ln|x^2-6x+34|+arctan(1/5 (x-3))+C