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

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

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

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

x^{2} - 6 x + 8 + 3 ln x^{2} - 6 x + 8 + x - 3 + C

Solution steps

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

Complete the square

x^{2} - 6 x + 8 : (x - 3)^{2} - 1

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

Apply u-substitution

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

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

Expand

\frac{u + 3}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} + \frac{3}{u ^{2} - 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{3}{u ^{2} - 1} d u = 3 ln u^{2} - 1 + u

= u^{2} - 1 + 3 ln u^{2} - 1 + u

Substitute back

u = x - 3

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

Simplify

(x - 3)^{2} - 1 + 3 ln (x - 3)^{2} - 1 + x - 3 : x^{2} - 6 x + 8 + 3 ln x^{2} - 6 x + 8 + x - 3

= x^{2} - 6 x + 8 + 3 ln x^{2} - 6 x + 8 + x - 3

Add a constant to the solution

= x^{2} - 6 x + 8 + 3 ln x^{2} - 6 x + 8 + x - 3 + C

x^{^{'}} = x^{2}, x (0) = 1

n = 0 \sum \infty (\frac{1}{2})^{2 n + 1}

\frac{1}{x + \frac{1}{x}}

x \to + (- 0) lim (f (x))

\frac{d}{d x} (cos (2 x) - sin (x))

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