What is the integral of x/(x^2-4x+5) ?

The integral of x/(x^2-4x+5) is 1/2 ln|x^2-4x+5|+2arctan(x-2)+C

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integral of x/(x^2-4x+5)

\int \frac{x}{x ^{2} - 4 x + 5} d x

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

Solution steps

\int \frac{x}{x ^{2} - 4 x + 5} d x

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

f (x) = \frac{x ^{4}}{2}

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

\int \frac{y + 2}{y} d y

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{n n}

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x - 4})

What is the integral of x/(x^2-4x+5) ?
The integral of x/(x^2-4x+5) is 1/2 ln|x^2-4x+5|+2arctan(x-2)+C