What is the integral of ((x+1))/(x^2-4x+12) ?

The integral of ((x+1))/(x^2-4x+12) is 1/2 ln|x^2-4x+12|+3/(2sqrt(2))arctan(1/(2sqrt(2))(x-2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x+1))/(x^2-4x+12)

\int \frac{( x + 1 )}{x ^{2} - 4 x + 12} d x

\frac{1}{2} ln x^{2} - 4 x + 12 + \frac{3}{2 2} arctan (\frac{1}{2 2} (x - 2)) + C

Solution steps

\int \frac{x + 1}{x ^{2} - 4 x + 12} d x

Complete the square

x^{2} - 4 x + 12 : (x - 2)^{2} + 8

= \int \frac{x + 1}{( x - 2 ) ^{2} + 8} d x

Apply u-substitution

: \int \frac{u + 3}{u ^{2} + 8} d u

= \int \frac{u + 3}{u ^{2} + 8} d u

Expand

\frac{u + 3}{u ^{2} + 8} : \frac{u}{u ^{2} + 8} + \frac{3}{u ^{2} + 8}

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} + 8} d u + \int \frac{3}{u ^{2} + 8} d u

\int \frac{u}{u ^{2} + 8} d u = \frac{1}{2} ln u^{2} + 8

\int \frac{3}{u ^{2} + 8} d u = \frac{3}{2 2} arctan (\frac{u}{2 2})

= \frac{1}{2} ln u^{2} + 8 + \frac{3}{2 2} arctan (\frac{u}{2 2})

Substitute back

u = x - 2

= \frac{1}{2} ln (x - 2)^{2} + 8 + \frac{3}{2 2} arctan (\frac{x - 2}{2 2})

Simplify

\frac{1}{2} ln (x - 2)^{2} + 8 + \frac{3}{2 2} arctan (\frac{x - 2}{2 2}) : \frac{1}{2} ln x^{2} - 4 x + 12 + \frac{3}{2 2} arctan (\frac{1}{2 2} (x - 2))

= \frac{1}{2} ln x^{2} - 4 x + 12 + \frac{3}{2 2} arctan (\frac{1}{2 2} (x - 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 4 x + 12 + \frac{3}{2 2} arctan (\frac{1}{2 2} (x - 2)) + C

f (x) = cos (3 x)

\frac{d}{d x} (3 + 2 sin (x))

\int_{0}^{x} a^{2} - x^{2} d x

\frac{d}{d x} (y^{2} - x^{3} + 8)

ln (x + 1)

What is the integral of ((x+1))/(x^2-4x+12) ?
The integral of ((x+1))/(x^2-4x+12) is 1/2 ln|x^2-4x+12|+3/(2sqrt(2))arctan(1/(2sqrt(2))(x-2))+C