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

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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

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{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

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

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

x^{^{'}} = x^{3} (t)

t a y l or x^{3} + 2 x + 1

d er i v a t i v e f (x) = \frac{5}{x ^{2}}

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