What is the integral of (2s-4)/((s^2-4s+20)^2) ?

The integral of (2s-4)/((s^2-4s+20)^2) is -1/(s^2-4s+20)+C

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integral of (2s-4)/((s^2-4s+20)^2)

\int \frac{2 s - 4}{( s ^{2} - 4 s + 20 ) ^{2}} d s

- \frac{1}{s ^{2} - 4 s + 20} + C

Solution steps

\int \frac{2 s - 4}{( s ^{2} - 4 s + 20 ) ^{2}} d s

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 = s^{2} - 4 s + 20

= - \frac{1}{s ^{2} - 4 s + 20}

Add a constant to the solution

= - \frac{1}{s ^{2} - 4 s + 20} + C

f (x) = \frac{x - 5}{x}

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

\int t^{9} e^{- t^{10}} d t

\int_{1}^{4} (2 - x^{2}) - (6 - 5 x) d x

\frac{d}{d x} (\frac{3 x + 4}{7 x - 5})

What is the integral of (2s-4)/((s^2-4s+20)^2) ?
The integral of (2s-4)/((s^2-4s+20)^2) is -1/(s^2-4s+20)+C