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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

= \int \frac{u + 3}{u ^{2} + 4} d u

Expand

\frac{u + 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} + \frac{3}{u ^{2} + 4}

= \int \frac{u}{u ^{2} + 4} + \frac{3}{u ^{2} + 4} 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{u}{u ^{2} + 4} d u + \int \frac{3}{u ^{2} + 4} d u

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

\int \frac{3}{u ^{2} + 4} d u = 3 ln (\frac{1}{2} u + 4 + u^{2})

= u^{2} + 4 + 3 ln (\frac{1}{2} u + 4 + u^{2})

Substitute back

u = x - 3

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

Simplify

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

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

Add a constant to the solution

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

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