What is the integral of (t-1)/(t^4+6t^3+9t^2) ?

The integral of (t-1)/(t^4+6t^3+9t^2) is 5/27 ln|t|-5/27 ln|t+3|+1/(3(t+3))+1/(9t)+1/(9(t+3))+C

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integral of (t-1)/(t^4+6t^3+9t^2)

\int \frac{t - 1}{t ^{4} + 6 t ^{3} + 9 t ^{2}} d t

\frac{5}{27} ln ∣ t ∣ - \frac{5}{27} ln ∣ t + 3 ∣ + \frac{1}{3 ( t + 3 )} + \frac{1}{9 t} + \frac{1}{9 ( t + 3 )} + C

Solution steps

\int \frac{t - 1}{t ^{4} + 6 t ^{3} + 9 t ^{2}} d t

Expand

\frac{t - 1}{t ^{4} + 6 t ^{3} + 9 t ^{2}} : \frac{t}{t ^{4} + 6 t ^{3} + 9 t ^{2}} - \frac{1}{t ^{4} + 6 t ^{3} + 9 t ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{t}{t ^{4} + 6 t ^{3} + 9 t ^{2}} d t - \int \frac{1}{t ^{4} + 6 t ^{3} + 9 t ^{2}} d t

\int \frac{t}{t ^{4} + 6 t ^{3} + 9 t ^{2}} d t = \frac{1}{9} ln ∣ t ∣ - \frac{1}{9} ln ∣ t + 3 ∣ + \frac{1}{3 ( t + 3 )}

\int \frac{1}{t ^{4} + 6 t ^{3} + 9 t ^{2}} d t = - \frac{2}{27} ln ∣ t ∣ - \frac{1}{9 t} + \frac{2}{27} ln ∣ t + 3 ∣ - \frac{1}{9 ( t + 3 )}

= \frac{1}{9} ln ∣ t ∣ - \frac{1}{9} ln ∣ t + 3 ∣ + \frac{1}{3 ( t + 3 )} - (- \frac{2}{27} ln ∣ t ∣ - \frac{1}{9 t} + \frac{2}{27} ln ∣ t + 3 ∣ - \frac{1}{9 ( t + 3 )})

Simplify

\frac{1}{9} ln ∣ t ∣ - \frac{1}{9} ln ∣ t + 3 ∣ + \frac{1}{3 ( t + 3 )} - (- \frac{2}{27} ln ∣ t ∣ - \frac{1}{9 t} + \frac{2}{27} ln ∣ t + 3 ∣ - \frac{1}{9 ( t + 3 )}) : \frac{5}{27} ln ∣ t ∣ - \frac{5}{27} ln ∣ t + 3 ∣ + \frac{1}{3 ( t + 3 )} + \frac{1}{9 t} + \frac{1}{9 ( t + 3 )}

= \frac{5}{27} ln ∣ t ∣ - \frac{5}{27} ln ∣ t + 3 ∣ + \frac{1}{3 ( t + 3 )} + \frac{1}{9 t} + \frac{1}{9 ( t + 3 )}

Add a constant to the solution

= \frac{5}{27} ln ∣ t ∣ - \frac{5}{27} ln ∣ t + 3 ∣ + \frac{1}{3 ( t + 3 )} + \frac{1}{9 t} + \frac{1}{9 ( t + 3 )} + C

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What is the integral of (t-1)/(t^4+6t^3+9t^2) ?
The integral of (t-1)/(t^4+6t^3+9t^2) is 5/27 ln|t|-5/27 ln|t+3|+1/(3(t+3))+1/(9t)+1/(9(t+3))+C