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

The integral of (t^2+2t-1)/(2t^3+3t^2-2t) is 1/2 ln|t|+1/10 ln|2t-1|-1/10 ln|t+2|+C

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

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

\frac{1}{2} ln ∣ t ∣ + \frac{1}{10} ln ∣ 2 t - 1 ∣ - \frac{1}{10} ln ∣ t + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{t ^{2} + 2 t - 1}{2 t ^{3} + 3 t ^{2} - 2 t} : \frac{1}{2 t} + \frac{1}{5 ( 2 t - 1 )} - \frac{1}{10 ( t + 2 )}

= \int \frac{1}{2 t} + \frac{1}{5 ( 2 t - 1 )} - \frac{1}{10 ( t + 2 )} 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

= \int \frac{1}{2 t} d t + \int \frac{1}{5 ( 2 t - 1 )} d t - \int \frac{1}{10 ( t + 2 )} d t

\int \frac{1}{2 t} d t = \frac{1}{2} ln ∣ t ∣

\int \frac{1}{5 ( 2 t - 1 )} d t = \frac{1}{10} ln ∣ 2 t - 1 ∣

\int \frac{1}{10 ( t + 2 )} d t = \frac{1}{10} ln ∣ t + 2 ∣

= \frac{1}{2} ln ∣ t ∣ + \frac{1}{10} ln ∣ 2 t - 1 ∣ - \frac{1}{10} ln ∣ t + 2 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ t ∣ + \frac{1}{10} ln ∣ 2 t - 1 ∣ - \frac{1}{10} ln ∣ t + 2 ∣ + C

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What is the integral of (t^2+2t-1)/(2t^3+3t^2-2t) ?
The integral of (t^2+2t-1)/(2t^3+3t^2-2t) is 1/2 ln|t|+1/10 ln|2t-1|-1/10 ln|t+2|+C