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

The integral of (x^2+2)^4 is (x^9)/9+(8x^7)/7+(24x^5)/5+(32x^3)/3+16x+C

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integral of (x^2+2)^4

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

\frac{x ^{9}}{9} + \frac{8 x ^{7}}{7} + \frac{24 x ^{5}}{5} + \frac{32 x ^{3}}{3} + 16 x + C

Solution steps

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

Expand

(x^{2} + 2)^{4} : x^{8} + 8 x^{6} + 24 x^{4} + 32 x^{2} + 16

= \int x^{8} + 8 x^{6} + 24 x^{4} + 32 x^{2} + 16 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^{8} d x + \int 8 x^{6} d x + \int 24 x^{4} d x + \int 32 x^{2} d x + \int 16 d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 8 x^{6} d x = \frac{8 x ^{7}}{7}

\int 24 x^{4} d x = \frac{24 x ^{5}}{5}

\int 32 x^{2} d x = \frac{32 x ^{3}}{3}

\int 16 d x = 16 x

= \frac{x ^{9}}{9} + \frac{8 x ^{7}}{7} + \frac{24 x ^{5}}{5} + \frac{32 x ^{3}}{3} + 16 x

Add a constant to the solution

= \frac{x ^{9}}{9} + \frac{8 x ^{7}}{7} + \frac{24 x ^{5}}{5} + \frac{32 x ^{3}}{3} + 16 x + C

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What is the integral of (x^2+2)^4 ?
The integral of (x^2+2)^4 is (x^9)/9+(8x^7)/7+(24x^5)/5+(32x^3)/3+16x+C