What is the integral of t^3+3t^2-2t+t^{-1} ?

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

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integral of t^3+3t^2-2t+t^{-1}

integral

t^{3} + 3 t^{2} - 2 t + t^{- 1}

\frac{t ^{4}}{4} + t^{3} - t^{2} + ln ∣ t ∣ + C

Solution steps

\int t^{3} + 3 t^{2} - 2 t + 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

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

\int t^{3} d t = \frac{t ^{4}}{4}

\int 3 t^{2} d t = t^{3}

\int 2 t d t = t^{2}

\int t^{- 1} d t = ln ∣ t ∣

= \frac{t ^{4}}{4} + t^{3} - t^{2} + ln ∣ t ∣

Add a constant to the solution

= \frac{t ^{4}}{4} + t^{3} - t^{2} + ln ∣ t ∣ + C

\frac{1}{tan ( x )}

\frac{x}{81 x ^{4} - 1}

\frac{2 x ^{2}}{( x ^{2} - 1 ) ^{2}}

\frac{1}{( 4 - x ^{2} )}

\frac{1}{( x ^{2} + x + 1 )}

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