What is the integral of (t^2)/(60)(t^3-5)^4+19 ?

The integral of (t^2)/(60)(t^3-5)^4+19 is 1/900 (t^3-5)^5+19t+C

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integral of (t^2)/(60)(t^3-5)^4+19

\int \frac{t ^{2}}{60} (t^{3} - 5)^{4} + 19 d t

\frac{1}{900} (t^{3} - 5)^{5} + 19 t + C

Solution steps

\int \frac{t ^{2}}{60} (t^{3} - 5)^{4} + 19 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 ^{2}}{60} (t^{3} - 5)^{4} d t + \int 19 d t

\int \frac{t ^{2}}{60} (t^{3} - 5)^{4} d t = \frac{1}{900} (t^{3} - 5)^{5}

\int 19 d t = 19 t

= \frac{1}{900} (t^{3} - 5)^{5} + 19 t

Add a constant to the solution

= \frac{1}{900} (t^{3} - 5)^{5} + 19 t + C

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What is the integral of (t^2)/(60)(t^3-5)^4+19 ?
The integral of (t^2)/(60)(t^3-5)^4+19 is 1/900 (t^3-5)^5+19t+C