What is the integral of 1/(sqrt(x^2-10x+34)) ?

The integral of 1/(sqrt(x^2-10x+34)) is ln(1/3 |x-5+sqrt(x^2-10x+34)|)+C

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integral of 1/(sqrt(x^2-10x+34))

\int \frac{1}{x ^{2} - 10 x + 34} d x

ln (\frac{1}{3} x - 5 + x^{2} - 10 x + 34) + C

Solution steps

\int \frac{1}{x ^{2} - 10 x + 34} d x

Complete the square

x^{2} - 10 x + 34 : (x - 5)^{2} + 9

= \int \frac{1}{( x - 5 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \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}{3} (x - 5))) + sec (arctan (\frac{1}{3} (x - 5)))

Simplify

ln tan (arctan (\frac{1}{3} (x - 5))) + sec (arctan (\frac{1}{3} (x - 5))) : ln (\frac{1}{3} x - 5 + x^{2} - 10 x + 34)

= ln (\frac{1}{3} x - 5 + x^{2} - 10 x + 34)

Add a constant to the solution

= ln (\frac{1}{3} x - 5 + x^{2} - 10 x + 34) + C

1 - 7 x^{2} y^{^{'}} = x

x \to 8 + lim (\frac{1}{x - 8})

\frac{\partial}{\partial x} (\frac{16}{x y})

\int_{0}^{2} \frac{5}{( x - 1 ) ^{\frac{1}{3}}} d x

x \to 2 lim (\frac{x ^{2} + 5}{x - 2})

What is the integral of 1/(sqrt(x^2-10x+34)) ?
The integral of 1/(sqrt(x^2-10x+34)) is ln(1/3 |x-5+sqrt(x^2-10x+34)|)+C