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

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

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

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

ln ∣ t - 1 ∣ - \frac{2}{t - 1} - ln ∣ t + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 4 \cdot \int \frac{1}{4 ( t - 1 )} + \frac{1}{2 ( t - 1 ) ^{2}} - \frac{1}{4 ( t + 1 )} d t

Apply the Sum Rule:

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

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

\int \frac{1}{4 ( t - 1 )} d t = \frac{1}{4} ln ∣ t - 1 ∣

\int \frac{1}{2 ( t - 1 ) ^{2}} d t = - \frac{1}{2 ( t - 1 )}

\int \frac{1}{4 ( t + 1 )} d t = \frac{1}{4} ln ∣ t + 1 ∣

= 4 (\frac{1}{4} ln ∣ t - 1 ∣ - \frac{1}{2 ( t - 1 )} - \frac{1}{4} ln ∣ t + 1 ∣)

Simplify

4 (\frac{1}{4} ln ∣ t - 1 ∣ - \frac{1}{2 ( t - 1 )} - \frac{1}{4} ln ∣ t + 1 ∣) : ln ∣ t - 1 ∣ - \frac{2}{t - 1} - ln ∣ t + 1 ∣

= ln ∣ t - 1 ∣ - \frac{2}{t - 1} - ln ∣ t + 1 ∣

Add a constant to the solution

= ln ∣ t - 1 ∣ - \frac{2}{t - 1} - ln ∣ t + 1 ∣ + C

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