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

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

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

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

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

Solution steps

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

Complete the square

x^{2} - 6 x : (x - 3)^{2} - 9

= \int \frac{x}{( x - 3 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{u + 3}{u ^{2} - 9} d u

= \int \frac{u + 3}{u ^{2} - 9} d u

Expand

\frac{u + 3}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} + \frac{3}{u ^{2} - 9}

= \int \frac{u}{u ^{2} - 9} + \frac{3}{u ^{2} - 9} 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} - 9} d u + \int \frac{3}{u ^{2} - 9} d u

\int \frac{u}{u ^{2} - 9} d u = u^{2} - 9

\int \frac{3}{u ^{2} - 9} d u = 3 ln (\frac{1}{3} u^{2} - 9 + u)

= u^{2} - 9 + 3 ln (\frac{1}{3} u^{2} - 9 + u)

Substitute back

u = x - 3

= (x - 3)^{2} - 9 + 3 ln (\frac{1}{3} (x - 3)^{2} - 9 + x - 3)

Simplify

(x - 3)^{2} - 9 + 3 ln (\frac{1}{3} (x - 3)^{2} - 9 + x - 3) : x^{2} - 6 x + 3 ln (\frac{1}{3} x^{2} - 6 x + x - 3)

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

Add a constant to the solution

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

y^{^{'}} + y = \frac{1}{1 + e ^{t}}

y^{^{'}} + 7 x^{6} y = x^{6}

x \to 0 lim (x^{2})

x \to 1 lim (6 x)

y = - 4 - 7 x^{2}, (- 4, - 16)

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