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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

y^{^{'}} = y^{2} - 3 y

\int x (2 x + 5)^{8} d x

\int e^{x} 4 - e^{2 x} d x

ln (sin (t))

\int \frac{x cos ( x )}{sin ^{2} ( x )} d x

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