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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \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}

= \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

= \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})

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x^{3} y e^{x y})

y = \frac{7}{x ^{2} + 3}

\frac{\partial}{\partial x} (- y - \frac{x}{y ^{2}})

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

\int e^{- x^{4}} (- 4 x^{3}) d x

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