What is the integral of (4x+1)/(x^2-6x+13) ?

The integral of (4x+1)/(x^2-6x+13) is 2ln|x^2-6x+13|+13/2 arctan(1/2 (x-3))+C

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integral of (4x+1)/(x^2-6x+13)

\int \frac{4 x + 1}{x ^{2} - 6 x + 13} d x

2 ln x^{2} - 6 x + 13 + \frac{13}{2} arctan (\frac{1}{2} (x - 3)) + C

Solution steps

\int \frac{4 x + 1}{x ^{2} - 6 x + 13} d x

Complete the square

x^{2} - 6 x + 13 : (x - 3)^{2} + 4

= \int \frac{4 x + 1}{( x - 3 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{4 u + 13}{u ^{2} + 4} d u

= \int \frac{4 u + 13}{u ^{2} + 4} d u

Expand

\frac{4 u + 13}{u ^{2} + 4} : \frac{4 u}{u ^{2} + 4} + \frac{13}{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{4 u}{u ^{2} + 4} d u + \int \frac{13}{u ^{2} + 4} d u

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

\int \frac{13}{u ^{2} + 4} d u = \frac{13}{2} arctan (\frac{u}{2})

= 2 ln u^{2} + 4 + \frac{13}{2} arctan (\frac{u}{2})

Substitute back

u = x - 3

= 2 ln (x - 3)^{2} + 4 + \frac{13}{2} arctan (\frac{x - 3}{2})

Simplify

2 ln (x - 3)^{2} + 4 + \frac{13}{2} arctan (\frac{x - 3}{2}) : 2 ln x^{2} - 6 x + 13 + \frac{13}{2} arctan (\frac{1}{2} (x - 3))

= 2 ln x^{2} - 6 x + 13 + \frac{13}{2} arctan (\frac{1}{2} (x - 3))

Add a constant to the solution

= 2 ln x^{2} - 6 x + 13 + \frac{13}{2} arctan (\frac{1}{2} (x - 3)) + C

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What is the integral of (4x+1)/(x^2-6x+13) ?
The integral of (4x+1)/(x^2-6x+13) is 2ln|x^2-6x+13|+13/2 arctan(1/2 (x-3))+C