What is the integral of (3x-3)/((x^2-2x+6)^2) ?

The integral of (3x-3)/((x^2-2x+6)^2) is -3/(2(x^2-2x+6))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x-3)/((x^2-2x+6)^2)

\int \frac{3 x - 3}{( x ^{2} - 2 x + 6 ) ^{2}} d x

- \frac{3}{2 ( x ^{2} - 2 x + 6 )} + C

Solution steps

\int \frac{3 x - 3}{( x ^{2} - 2 x + 6 ) ^{2}} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{3}{2} (- \frac{1}{u})

Substitute back

u = x^{2} - 2 x + 6

= \frac{3}{2} (- \frac{1}{x ^{2} - 2 x + 6})

Simplify

\frac{3}{2} (- \frac{1}{x ^{2} - 2 x + 6}) : - \frac{3}{2 ( x ^{2} - 2 x + 6 )}

= - \frac{3}{2 ( x ^{2} - 2 x + 6 )}

Add a constant to the solution

= - \frac{3}{2 ( x ^{2} - 2 x + 6 )} + C

\int 4 e^{2 x} d x

\frac{d}{d x} (\frac{1}{x ^{2} - 9})

\frac{d}{d x} ((x + e^{x}) (3 - x))

\int \frac{x ^{2}}{9 - x} d x

\frac{5 ( 1 - sin ( x ) )}{2 cos ( x )}

What is the integral of (3x-3)/((x^2-2x+6)^2) ?
The integral of (3x-3)/((x^2-2x+6)^2) is -3/(2(x^2-2x+6))+C