What is the integral of ((t-2))/((t^2-4t+3)^3) ?

The integral of ((t-2))/((t^2-4t+3)^3) is -1/(4(t^2-4t+3)^2)+C

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integral of ((t-2))/((t^2-4t+3)^3)

\int \frac{( t - 2 )}{( t ^{2} - 4 t + 3 ) ^{3}} d t

- \frac{1}{4 ( t ^{2} - 4 t + 3 ) ^{2}} + C

Solution steps

\int \frac{t - 2}{( t ^{2} - 4 t + 3 ) ^{3}} d t

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= \int \frac{1}{2 u ^{3}} 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 ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = t^{2} - 4 t + 3

= \frac{1}{2} (- \frac{1}{2 ( t ^{2} - 4 t + 3 ) ^{2}})

Simplify

\frac{1}{2} (- \frac{1}{2 ( t ^{2} - 4 t + 3 ) ^{2}}) : - \frac{1}{4 ( t ^{2} - 4 t + 3 ) ^{2}}

= - \frac{1}{4 ( t ^{2} - 4 t + 3 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( t ^{2} - 4 t + 3 ) ^{2}} + C

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What is the integral of ((t-2))/((t^2-4t+3)^3) ?
The integral of ((t-2))/((t^2-4t+3)^3) is -1/(4(t^2-4t+3)^2)+C