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

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

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

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

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

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (arccot (\frac{x}{4}))

\int \frac{y ^{3}}{( 2 + 3 y ^{4} ) ^{3}} d y

sin (tan (3 x))

lo g_{4} (e^{2 x} cos (3 x))

\frac{30 x ^{2}}{( - 5 x ^{3} + 4 ) ^{3}}

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