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

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

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integral of (2x-1)/((2x^2-2x+4)^2)

\int \frac{2 x - 1}{( 2 x ^{2} - 2 x + 4 ) ^{2}} d x

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

Solution steps

\int \frac{2 x - 1}{( 2 x ^{2} - 2 x + 4 ) ^{2}} d x

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = 2 x^{2} - 2 x + 4

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

Simplify

\frac{1}{2} (- \frac{1}{2 x ^{2} - 2 x + 4}) : - \frac{1}{2 ( 2 x ^{2} - 2 x + 4 )}

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

Add a constant to the solution

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

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