What is the integral of (2x)/(sqrt(x^2-2x+5)) ?

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

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integral of (2x)/(sqrt(x^2-2x+5))

\int \frac{2 x}{x ^{2} - 2 x + 5} d x

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

Solution steps

\int \frac{2 x}{x ^{2} - 2 x + 5} d x

Take the constant out:

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

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

Complete the square

x^{2} - 2 x + 5 : (x - 1)^{2} + 4

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

Apply u-substitution

= 2 \cdot \int \frac{u + 1}{u ^{2} + 4} d u

Expand

\frac{u + 1}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} + \frac{1}{u ^{2} + 4}

= 2 \cdot \int \frac{u}{u ^{2} + 4} + \frac{1}{u ^{2} + 4} 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

= 2 (\int \frac{u}{u ^{2} + 4} d u + \int \frac{1}{u ^{2} + 4} d u)

\int \frac{u}{u ^{2} + 4} d u = u^{2} + 4

\int \frac{1}{u ^{2} + 4} d u = ln (\frac{1}{2} u + 4 + u^{2})

= 2 (u^{2} + 4 + ln (\frac{1}{2} u + 4 + u^{2}))

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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