What is the integral of 1/(sqrt(x^2-6x+10)) ?

The integral of 1/(sqrt(x^2-6x+10)) is ln|sqrt(x^2-6x+10)+x-3|+C

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integral of 1/(sqrt(x^2-6x+10))

\int \frac{1}{x ^{2} - 6 x + 10} d x

ln x^{2} - 6 x + 10 + x - 3 + C

Solution steps

\int \frac{1}{x ^{2} - 6 x + 10} d x

Complete the square

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

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

Apply u-substitution

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

Use the common integral:

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

= ln u^{2} + 1 + u

Substitute back

u = x - 3

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

Expand

(x - 3)^{2} + 1 : x^{2} - 6 x + 10

= ln x^{2} - 6 x + 10 + x - 3

Add a constant to the solution

= ln x^{2} - 6 x + 10 + x - 3 + C

What is the integral of 1/(sqrt(x^2-6x+10)) ?
The integral of 1/(sqrt(x^2-6x+10)) is ln|sqrt(x^2-6x+10)+x-3|+C