What is the integral of x^5(x^2+3)^4 ?

The integral of x^5(x^2+3)^4 is (x^{14}}{14}+x^{12}+\frac{27x^{10})/5+(27x^8)/2+(27x^6)/2+C

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integral of x^5(x^2+3)^4

integral

x^{5} (x^{2} + 3)^{4}

\frac{x ^{14}}{14} + x^{12} + \frac{27 x ^{10}}{5} + \frac{27 x ^{8}}{2} + \frac{27 x ^{6}}{2} + C

Solution steps

\int x^{5} (x^{2} + 3)^{4} d x

Expand

x^{5} (x^{2} + 3)^{4} : x^{13} + 12 x^{11} + 54 x^{9} + 108 x^{7} + 81 x^{5}

= \int x^{13} + 12 x^{11} + 54 x^{9} + 108 x^{7} + 81 x^{5} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{13} d x + \int 12 x^{11} d x + \int 54 x^{9} d x + \int 108 x^{7} d x + \int 81 x^{5} d x

\int x^{13} d x = \frac{x ^{14}}{14}

\int 12 x^{11} d x = x^{12}

\int 54 x^{9} d x = \frac{27 x ^{10}}{5}

\int 108 x^{7} d x = \frac{27 x ^{8}}{2}

\int 81 x^{5} d x = \frac{27 x ^{6}}{2}

= \frac{x ^{14}}{14} + x^{12} + \frac{27 x ^{10}}{5} + \frac{27 x ^{8}}{2} + \frac{27 x ^{6}}{2}

Add a constant to the solution

= \frac{x ^{14}}{14} + x^{12} + \frac{27 x ^{10}}{5} + \frac{27 x ^{8}}{2} + \frac{27 x ^{6}}{2} + C

- 3 x^{- 4}

cos (x^{2} + 1)

(4 + x)^{\frac{1}{2}}

x^{2} + 2 x + 3

\frac{s}{( s ^{2} + 1 )}

What is the integral of x^5(x^2+3)^4 ?
The integral of x^5(x^2+3)^4 is (x^{14}}{14}+x^{12}+\frac{27x^{10})/5+(27x^8)/2+(27x^6)/2+C