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

The integral of ((t^4)/2-3/(t^{-2)}+3t) is (t^5}{10}-t^3+\frac{3t^2)/2+C

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

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

\frac{t ^{5}}{10} - t^{3} + \frac{3 t ^{2}}{2} + C

Solution steps

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

\int \frac{t ^{4}}{2} d t = \frac{t ^{5}}{10}

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

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

= \frac{t ^{5}}{10} - t^{3} + \frac{3 t ^{2}}{2}

Add a constant to the solution

= \frac{t ^{5}}{10} - t^{3} + \frac{3 t ^{2}}{2} + C

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What is the integral of ((t^4)/2-3/(t^{-2)}+3t) ?
The integral of ((t^4)/2-3/(t^{-2)}+3t) is (t^5}{10}-t^3+\frac{3t^2)/2+C