What is the integral of 2/(sqrt(x^2-4x+9)) ?

The integral of 2/(sqrt(x^2-4x+9)) is 2ln| 1/(sqrt(5))(x-2)+sqrt(1/5 (x^2-4x+9))|+C

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integral of 2/(sqrt(x^2-4x+9))

\int \frac{2}{x ^{2} - 4 x + 9} d x

2 ln \frac{1}{5} (x - 2) + \frac{1}{5} (x^{2} - 4 x + 9) + C

Solution steps

\int \frac{2}{x ^{2} - 4 x + 9} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{2} - 4 x + 9} d x

Complete the square

x^{2} - 4 x + 9 : (x - 2)^{2} + 5

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

Apply u-substitution

= 2 \cdot \int \frac{1}{u ^{2} + 5} d u

Apply Trigonometric Substitution

= 2 \cdot \int sec (v) d v

Use the common integral:

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

= 2 ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

2 ln tan (arctan (\frac{1}{5} (x - 2))) + sec (arctan (\frac{1}{5} (x - 2))) : 2 ln \frac{1}{5} (x - 2) + \frac{1}{5} (x^{2} - 4 x + 9)

= 2 ln \frac{1}{5} (x - 2) + \frac{1}{5} (x^{2} - 4 x + 9)

Add a constant to the solution

= 2 ln \frac{1}{5} (x - 2) + \frac{1}{5} (x^{2} - 4 x + 9) + C

What is the integral of 2/(sqrt(x^2-4x+9)) ?
The integral of 2/(sqrt(x^2-4x+9)) is 2ln| 1/(sqrt(5))(x-2)+sqrt(1/5 (x^2-4x+9))|+C