What is the integral of 9x^4(x^5+2)^8 ?

The integral of 9x^4(x^5+2)^8 is

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integral of 9x^4(x^5+2)^8

\int 9 x^{4} (x^{5} + 2)^{8} d x

9 (\frac{x ^{45}}{45} + \frac{2 x ^{40}}{5} + \frac{16 x ^{35}}{5} + \frac{224 x ^{30}}{15} + \frac{224 x ^{25}}{5} + \frac{448 x ^{20}}{5} + \frac{1792 x ^{15}}{15} + \frac{512 x ^{10}}{5} + \frac{256 x ^{5}}{5}) + C

Solution steps

\int 9 x^{4} (x^{5} + 2)^{8} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int x^{4} (x^{5} + 2)^{8} d x

Expand

x^{4} (x^{5} + 2)^{8} : x^{44} + 16 x^{39} + 112 x^{34} + 448 x^{29} + 1120 x^{24} + 1792 x^{19} + 1792 x^{14} + 1024 x^{9} + 256 x^{4}

= 9 \cdot \int x^{44} + 16 x^{39} + 112 x^{34} + 448 x^{29} + 1120 x^{24} + 1792 x^{19} + 1792 x^{14} + 1024 x^{9} + 256 x^{4} 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

= 9 (\int x^{44} d x + \int 16 x^{39} d x + \int 112 x^{34} d x + \int 448 x^{29} d x + \int 1120 x^{24} d x + \int 1792 x^{19} d x + \int 1792 x^{14} d x + \int 1024 x^{9} d x + \int 256 x^{4} d x)

\int x^{44} d x = \frac{x ^{45}}{45}

\int 16 x^{39} d x = \frac{2 x ^{40}}{5}

\int 112 x^{34} d x = \frac{16 x ^{35}}{5}

\int 448 x^{29} d x = \frac{224 x ^{30}}{15}

\int 1120 x^{24} d x = \frac{224 x ^{25}}{5}

\int 1792 x^{19} d x = \frac{448 x ^{20}}{5}

\int 1792 x^{14} d x = \frac{1792 x ^{15}}{15}

\int 1024 x^{9} d x = \frac{512 x ^{10}}{5}

\int 256 x^{4} d x = \frac{256 x ^{5}}{5}

= 9 (\frac{x ^{45}}{45} + \frac{2 x ^{40}}{5} + \frac{16 x ^{35}}{5} + \frac{224 x ^{30}}{15} + \frac{224 x ^{25}}{5} + \frac{448 x ^{20}}{5} + \frac{1792 x ^{15}}{15} + \frac{512 x ^{10}}{5} + \frac{256 x ^{5}}{5})

Add a constant to the solution

= 9 (\frac{x ^{45}}{45} + \frac{2 x ^{40}}{5} + \frac{16 x ^{35}}{5} + \frac{224 x ^{30}}{15} + \frac{224 x ^{25}}{5} + \frac{448 x ^{20}}{5} + \frac{1792 x ^{15}}{15} + \frac{512 x ^{10}}{5} + \frac{256 x ^{5}}{5}) + C