What is the integral of 8x(4x^2+1)^5 ?

The integral of 8x(4x^2+1)^5 is 8((256x^{12})/3+128x^{10}+80x^8+(80x^6)/3+5x^4+(x^2)/2)+C

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integral of 8x(4x^2+1)^5

\int 8 x (4 x^{2} + 1)^{5} d x

8 (\frac{256 x ^{12}}{3} + 128 x^{10} + 80 x^{8} + \frac{80 x ^{6}}{3} + 5 x^{4} + \frac{x ^{2}}{2}) + C

Solution steps

\int 8 x (4 x^{2} + 1)^{5} d x

Take the constant out:

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

= 8 \cdot \int x (4 x^{2} + 1)^{5} d x

Expand

x (4 x^{2} + 1)^{5} : 1024 x^{11} + 1280 x^{9} + 640 x^{7} + 160 x^{5} + 20 x^{3} + x

= 8 \cdot \int 1024 x^{11} + 1280 x^{9} + 640 x^{7} + 160 x^{5} + 20 x^{3} + x 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

= 8 (\int 1024 x^{11} d x + \int 1280 x^{9} d x + \int 640 x^{7} d x + \int 160 x^{5} d x + \int 20 x^{3} d x + \int x d x)

\int 1024 x^{11} d x = \frac{256 x ^{12}}{3}

\int 1280 x^{9} d x = 128 x^{10}

\int 640 x^{7} d x = 80 x^{8}

\int 160 x^{5} d x = \frac{80 x ^{6}}{3}

\int 20 x^{3} d x = 5 x^{4}

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

= 8 (\frac{256 x ^{12}}{3} + 128 x^{10} + 80 x^{8} + \frac{80 x ^{6}}{3} + 5 x^{4} + \frac{x ^{2}}{2})

Add a constant to the solution

= 8 (\frac{256 x ^{12}}{3} + 128 x^{10} + 80 x^{8} + \frac{80 x ^{6}}{3} + 5 x^{4} + \frac{x ^{2}}{2}) + C

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What is the integral of 8x(4x^2+1)^5 ?
The integral of 8x(4x^2+1)^5 is 8((256x^{12})/3+128x^{10}+80x^8+(80x^6)/3+5x^4+(x^2)/2)+C