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

The integral of (2x-3)/((x^2-3x+6)^2) is -1/(x^2-3x+6)+C

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

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

- \frac{1}{x ^{2} - 3 x + 6} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x^{2} - 3 x + 6

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

Add a constant to the solution

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

\int tan (x) - x d x

x \to 4 lim (\frac{x}{x ^{2} + 4})

arctan (3 x)

\int \frac{( 36 - 25 x ^{2} ) ^{\frac{3}{2}}}{x ^{6}} d x

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

What is the integral of (2x-3)/((x^2-3x+6)^2) ?
The integral of (2x-3)/((x^2-3x+6)^2) is -1/(x^2-3x+6)+C