What is the integral of (2x-4)/((x^2-4x+7)^2) ?

The integral of (2x-4)/((x^2-4x+7)^2) is -1/(x^2-4x+7)+C

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integral of (2x-4)/((x^2-4x+7)^2)

\int \frac{2 x - 4}{( x ^{2} - 4 x + 7 ) ^{2}} d x

- \frac{1}{x ^{2} - 4 x + 7} + C

Solution steps

\int \frac{2 x - 4}{( x ^{2} - 4 x + 7 ) ^{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} - 4 x + 7

= - \frac{1}{x ^{2} - 4 x + 7}

Add a constant to the solution

= - \frac{1}{x ^{2} - 4 x + 7} + C

(- 2)^{t - 1}

\int \frac{3 x ( e ^{- 2 x} - 7 x )}{5} d x

\int (arcsin (x))^{2} d x

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

What is the integral of (2x-4)/((x^2-4x+7)^2) ?
The integral of (2x-4)/((x^2-4x+7)^2) is -1/(x^2-4x+7)+C