What is the integral of (x^2)/(x^2-6x+13) ?

The integral of (x^2)/(x^2-6x+13) is 3ln|x^2-6x+13|+5/2 arctan(1/2 (x-3))+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(x^2-6x+13)

\int \frac{x ^{2}}{x ^{2} - 6 x + 13} d x

3 ln x^{2} - 6 x + 13 + \frac{5}{2} arctan (\frac{1}{2} (x - 3)) + x + C

Solution steps

\int \frac{x ^{2}}{x ^{2} - 6 x + 13} d x

\frac{x ^{2}}{x ^{2} - 6 x + 13} = \frac{6 x - 13}{x ^{2} - 6 x + 13} + 1

= \int \frac{6 x - 13}{x ^{2} - 6 x + 13} + 1 d x

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{6 x - 13}{x ^{2} - 6 x + 13} d x + \int 1 d x

\int \frac{6 x - 13}{x ^{2} - 6 x + 13} d x = 3 ln x^{2} - 6 x + 13 + \frac{5}{2} arctan (\frac{1}{2} (x - 3))

\int 1 d x = x

= 3 ln x^{2} - 6 x + 13 + \frac{5}{2} arctan (\frac{1}{2} (x - 3)) + x

Add a constant to the solution

= 3 ln x^{2} - 6 x + 13 + \frac{5}{2} arctan (\frac{1}{2} (x - 3)) + x + C

(1 + x^{4}) d y = - x (1 + 4 y^{2}) d x

x \to 0 lim (\frac{1}{x} \cdot sin (x))

y^{^{'}} = x y

\frac{d}{d x} (7 x^{2} - 12 x)

\frac{d}{d x} (\frac{3}{2} x - 5)

What is the integral of (x^2)/(x^2-6x+13) ?
The integral of (x^2)/(x^2-6x+13) is 3ln|x^2-6x+13|+5/2 arctan(1/2 (x-3))+x+C