What is the integral of ((6x-9))/((x^2-3x+5)^3) ?

The integral of ((6x-9))/((x^2-3x+5)^3) is -3/(2(x^2-3x+5)^2)+C

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integral of ((6x-9))/((x^2-3x+5)^3)

\int \frac{( 6 x - 9 )}{( x ^{2} - 3 x + 5 ) ^{3}} d x

- \frac{3}{2 ( x ^{2} - 3 x + 5 ) ^{2}} + C

Solution steps

\int \frac{6 x - 9}{( x ^{2} - 3 x + 5 ) ^{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = x^{2} - 3 x + 5

= 3 (- \frac{1}{2 ( x ^{2} - 3 x + 5 ) ^{2}})

Simplify

3 (- \frac{1}{2 ( x ^{2} - 3 x + 5 ) ^{2}}) : - \frac{3}{2 ( x ^{2} - 3 x + 5 ) ^{2}}

= - \frac{3}{2 ( x ^{2} - 3 x + 5 ) ^{2}}

Add a constant to the solution

= - \frac{3}{2 ( x ^{2} - 3 x + 5 ) ^{2}} + C

\frac{d}{d x} (- \frac{1}{3} cos (3 x))

y = ln (x^{5})

f (x) = (\frac{x + 1}{x - 1})^{7}

\frac{d}{d x} (lo g_{4} (cos (2 x)))

\frac{d}{d x} (- \frac{3}{x ^{2}} + 4 e^{2 x})

What is the integral of ((6x-9))/((x^2-3x+5)^3) ?
The integral of ((6x-9))/((x^2-3x+5)^3) is -3/(2(x^2-3x+5)^2)+C