What is the integral of x/(sqrt(x^2-2ax+2a^2)) ?

The integral of x/(sqrt(x^2-2ax+2a^2)) is sqrt(x^2-2ax+2a^2)+aln((|x-a+sqrt(x^2-2ax+2a^2)|)/(|a|))+C

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integral of x/(sqrt(x^2-2ax+2a^2))

\int \frac{x}{x ^{2} - 2 a x + 2 a ^{2}} d x

x^{2} - 2 a x + 2 a^{2} + a ln (\frac{x - a + x ^{2} - 2 a x + 2 a ^{2}}{∣ a ∣}) + C

Solution steps

\int \frac{x}{x ^{2} - 2 a x + 2 a ^{2}} d x

Complete the square

x^{2} - 2 a x + 2 a^{2} : (x - a)^{2} + a^{2}

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

Apply u-substitution

= \int \frac{u + a}{u ^{2} + a ^{2}} d u

Expand

\frac{u + a}{u ^{2} + a ^{2}} : \frac{u}{u ^{2} + a ^{2}} + \frac{a}{u ^{2} + a ^{2}}

= \int \frac{u}{u ^{2} + a ^{2}} + \frac{a}{u ^{2} + a ^{2}} 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} + a ^{2}} d u + \int \frac{a}{u ^{2} + a ^{2}} d u

\int \frac{u}{u ^{2} + a ^{2}} d u = u^{2} + a^{2}

\int \frac{a}{u ^{2} + a ^{2}} d u = a ln (\frac{u + a ^{2} + u ^{2}}{∣ a ∣})

= u^{2} + a^{2} + a ln (\frac{u + a ^{2} + u ^{2}}{∣ a ∣})

Substitute back

u = x - a

= (x - a)^{2} + a^{2} + a ln \frac{x - a + a ^{2} + ( x - a ) ^{2}}{∣ a ∣}

Simplify

(x - a)^{2} + a^{2} + a ln \frac{x - a + a ^{2} + ( x - a ) ^{2}}{∣ a ∣} : x^{2} - 2 a x + 2 a^{2} + a ln (\frac{x - a + x ^{2} - 2 a x + 2 a ^{2}}{∣ a ∣})

= x^{2} - 2 a x + 2 a^{2} + a ln (\frac{x - a + x ^{2} - 2 a x + 2 a ^{2}}{∣ a ∣})

Add a constant to the solution

= x^{2} - 2 a x + 2 a^{2} + a ln (\frac{x - a + x ^{2} - 2 a x + 2 a ^{2}}{∣ a ∣}) + C

What is the integral of x/(sqrt(x^2-2ax+2a^2)) ?
The integral of x/(sqrt(x^2-2ax+2a^2)) is sqrt(x^2-2ax+2a^2)+aln((|x-a+sqrt(x^2-2ax+2a^2)|)/(|a|))+C