What is the integral of (x^6+8)^2 ?

The integral of (x^6+8)^2 is (x^{13}}{13}+\frac{16x^7)/7+64x+C

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integral of (x^6+8)^2

\int (x^{6} + 8)^{2} d x

\frac{x ^{13}}{13} + \frac{16 x ^{7}}{7} + 64 x + C

Solution steps

\int (x^{6} + 8)^{2} d x

Expand

(x^{6} + 8)^{2} : x^{12} + 16 x^{6} + 64

= \int x^{12} + 16 x^{6} + 64 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^{12} d x + \int 16 x^{6} d x + \int 64 d x

\int x^{12} d x = \frac{x ^{13}}{13}

\int 16 x^{6} d x = \frac{16 x ^{7}}{7}

\int 64 d x = 64 x

= \frac{x ^{13}}{13} + \frac{16 x ^{7}}{7} + 64 x

Add a constant to the solution

= \frac{x ^{13}}{13} + \frac{16 x ^{7}}{7} + 64 x + C

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What is the integral of (x^6+8)^2 ?
The integral of (x^6+8)^2 is (x^{13}}{13}+\frac{16x^7)/7+64x+C