What is the integral of 3/(sqrt(x^2-8x+6)) ?

The integral of 3/(sqrt(x^2-8x+6)) is 3ln|sqrt(1/10 (x^2-8x+6))+1/(sqrt(10))(x-4)|+C

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integral of 3/(sqrt(x^2-8x+6))

\int \frac{3}{x ^{2} - 8 x + 6} d x

3 ln \frac{1}{10} (x^{2} - 8 x + 6) + \frac{1}{10} (x - 4) + C

Solution steps

\int \frac{3}{x ^{2} - 8 x + 6} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} - 8 x + 6} d x

Complete the square

x^{2} - 8 x + 6 : (x - 4)^{2} - 10

= 3 \cdot \int \frac{1}{( x - 4 ) ^{2} - 10} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 10} d u

= 3 \cdot \int \frac{1}{u ^{2} - 10} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= 3 \cdot \int sec (v) d v

Use the common integral:

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

= 3 ln ∣ tan (v) + sec (v) ∣

Substitute back

= 3 ln tan (arcsec (\frac{1}{10} (x - 4))) + sec (arcsec (\frac{1}{10} (x - 4)))

Simplify

3 ln tan (arcsec (\frac{1}{10} (x - 4))) + sec (arcsec (\frac{1}{10} (x - 4))) : 3 ln \frac{1}{10} (x^{2} - 8 x + 6) + \frac{1}{10} (x - 4)

= 3 ln \frac{1}{10} (x^{2} - 8 x + 6) + \frac{1}{10} (x - 4)

Add a constant to the solution

= 3 ln \frac{1}{10} (x^{2} - 8 x + 6) + \frac{1}{10} (x - 4) + C

k = 0 \sum \infty x^{k + 1}

y = (3 x + 8) x^{2}

x \to 2 - lim (x^{2} + 1)

\frac{d y}{d x}, x y = 9

\frac{d y}{7 a + y} = d x cot (x)

What is the integral of 3/(sqrt(x^2-8x+6)) ?
The integral of 3/(sqrt(x^2-8x+6)) is 3ln|sqrt(1/10 (x^2-8x+6))+1/(sqrt(10))(x-4)|+C