What is the integral from 0 to 2 of t^3(1+t^4)^3 ?

The integral from 0 to 2 of t^3(1+t^4)^3 is 5220

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integral from 0 to 2 of t^3(1+t^4)^3

\int_{0}^{2} t^{3} (1 + t^{4})^{3} d t

5220

Solution steps

\int_{0}^{2} t^{3} (1 + t^{4})^{3} d t

Expand

t^{3} (1 + t^{4})^{3} : t^{3} + 3 t^{7} + 3 t^{11} + t^{15}

= \int_{0}^{2} t^{3} + 3 t^{7} + 3 t^{11} + t^{15} 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_{0}^{2} t^{3} d t + \int_{0}^{2} 3 t^{7} d t + \int_{0}^{2} 3 t^{11} d t + \int_{0}^{2} t^{15} d t

\int_{0}^{2} t^{3} d t = 4

\int_{0}^{2} 3 t^{7} d t = 96

\int_{0}^{2} 3 t^{11} d t = 1024

\int_{0}^{2} t^{15} d t = 4096

= 4 + 96 + 1024 + 4096

Simplify

= 5220

f (x) = x, (- 1, 1)

\int \frac{1}{x ( 2 x + 5 )} d x

\int \frac{x + 2}{x ^{2} + 4 x + 5} d x

f (x) = \frac{x ^{3}}{2}, (5, 62.5)

\int \frac{e ^{x + 666}}{x} d x

What is the integral from 0 to 2 of t^3(1+t^4)^3 ?
The integral from 0 to 2 of t^3(1+t^4)^3 is 5220