What is the integral from 0 to x of (t^3+1)^{10} ?

The integral from 0 to x of (t^3+1)^{10} is

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integral from 0 to x of (t^3+1)^{10}

\int_{0}^{x} (t^{3} + 1)^{10} d t

\frac{x ^{31}}{31} + \frac{5 x ^{28}}{14} + \frac{9 x ^{25}}{5} + \frac{60 x ^{22}}{11} + \frac{210 x ^{19}}{19} + \frac{63 x ^{16}}{4} + \frac{210 x ^{13}}{13} + 12 x^{10} + \frac{45 x ^{7}}{7} + \frac{5 x ^{4}}{2} + x

Solution steps

\int_{0}^{x} (t^{3} + 1)^{10} d t

Expand

(t^{3} + 1)^{10} : t^{30} + 10 t^{27} + 45 t^{24} + 120 t^{21} + 210 t^{18} + 252 t^{15} + 210 t^{12} + 120 t^{9} + 45 t^{6} + 10 t^{3} + 1

= \int_{0}^{x} t^{30} + 10 t^{27} + 45 t^{24} + 120 t^{21} + 210 t^{18} + 252 t^{15} + 210 t^{12} + 120 t^{9} + 45 t^{6} + 10 t^{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_{0}^{x} t^{30} d t + \int_{0}^{x} 10 t^{27} d t + \int_{0}^{x} 45 t^{24} d t + \int_{0}^{x} 120 t^{21} d t + \int_{0}^{x} 210 t^{18} d t + \int_{0}^{x} 252 t^{15} d t + \int_{0}^{x} 210 t^{12} d t + \int_{0}^{x} 120 t^{9} d t + \int_{0}^{x} 45 t^{6} d t + \int_{0}^{x} 10 t^{3} d t + \int_{0}^{x} 1 d t

\int_{0}^{x} t^{30} d t = \frac{x ^{31}}{31}

\int_{0}^{x} 10 t^{27} d t = \frac{5 x ^{28}}{14}

\int_{0}^{x} 45 t^{24} d t = \frac{9 x ^{25}}{5}

\int_{0}^{x} 120 t^{21} d t = \frac{60 x ^{22}}{11}

\int_{0}^{x} 210 t^{18} d t = \frac{210 x ^{19}}{19}

\int_{0}^{x} 252 t^{15} d t = \frac{63 x ^{16}}{4}

\int_{0}^{x} 210 t^{12} d t = \frac{210 x ^{13}}{13}

\int_{0}^{x} 120 t^{9} d t = 12 x^{10}

\int_{0}^{x} 45 t^{6} d t = \frac{45 x ^{7}}{7}

\int_{0}^{x} 10 t^{3} d t = \frac{5 x ^{4}}{2}

\int_{0}^{x} 1 d t = x

= \frac{x ^{31}}{31} + \frac{5 x ^{28}}{14} + \frac{9 x ^{25}}{5} + \frac{60 x ^{22}}{11} + \frac{210 x ^{19}}{19} + \frac{63 x ^{16}}{4} + \frac{210 x ^{13}}{13} + 12 x^{10} + \frac{45 x ^{7}}{7} + \frac{5 x ^{4}}{2} + x

What is the integral from 0 to x of (t^3+1)^{10} ?
The integral from 0 to x of (t^3+1)^{10} is