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

The integral of (t^3)/3+1 is (t^4)/(12)+t+C

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integral of (t^3)/3+1

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

\frac{t ^{4}}{12} + t + C

Solution steps

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

\int \frac{t ^{3}}{3} d t = \frac{t ^{4}}{12}

\int 1 d t = t

= \frac{t ^{4}}{12} + t

Add a constant to the solution

= \frac{t ^{4}}{12} + t + C

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