What is the integral of 1/(sqrt(x^2-10x+29)) ?

The integral of 1/(sqrt(x^2-10x+29)) is ln(1/2 |x-5+sqrt(x^2-10x+29)|)+C

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integral of 1/(sqrt(x^2-10x+29))

\int \frac{1}{x ^{2} - 10 x + 29} d x

ln (\frac{1}{2} x - 5 + x^{2} - 10 x + 29) + C

Solution steps

\int \frac{1}{x ^{2} - 10 x + 29} d x

Complete the square

x^{2} - 10 x + 29 : (x - 5)^{2} + 4

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

Apply u-substitution

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

Apply Trigonometric Substitution

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{2} (x - 5))) + sec (arctan (\frac{1}{2} (x - 5)))

Simplify

ln tan (arctan (\frac{1}{2} (x - 5))) + sec (arctan (\frac{1}{2} (x - 5))) : ln (\frac{1}{2} x - 5 + x^{2} - 10 x + 29)

= ln (\frac{1}{2} x - 5 + x^{2} - 10 x + 29)

Add a constant to the solution

= ln (\frac{1}{2} x - 5 + x^{2} - 10 x + 29) + C

What is the integral of 1/(sqrt(x^2-10x+29)) ?
The integral of 1/(sqrt(x^2-10x+29)) is ln(1/2 |x-5+sqrt(x^2-10x+29)|)+C