What is the integral of 1/(sqrt(x^2-2x+4)) ?

The integral of 1/(sqrt(x^2-2x+4)) is ln| 1/(sqrt(3))(x-1)+sqrt(1/3 (x^2-2x+4))|+C

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integral of 1/(sqrt(x^2-2x+4))

\int \frac{1}{x ^{2} - 2 x + 4} d x

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

Solution steps

\int \frac{1}{x ^{2} - 2 x + 4} d x

Complete the square

x^{2} - 2 x + 4 : (x - 1)^{2} + 3

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 3} d u

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

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

Add a constant to the solution

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

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What is the integral of 1/(sqrt(x^2-2x+4)) ?
The integral of 1/(sqrt(x^2-2x+4)) is ln| 1/(sqrt(3))(x-1)+sqrt(1/3 (x^2-2x+4))|+C