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

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

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

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

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

Solution steps

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

Complete the square

x^{2} - 2 x + 10 : (x - 1)^{2} + 9

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

Simplify

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

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

Add a constant to the solution

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

49 y^{^{''}} + 28 y^{^{'}} + 4 y = 0

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 94

x \to 0 lim (x \cdot sin (\frac{π}{x}))

m = \frac{9}{5 + h} - \frac{9}{5}

n = 1 \sum \infty e^{- n} + 1

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