What is the integral of 8x(x^2+1)^3 ?

The integral of 8x(x^2+1)^3 is x^8+4x^6+6x^4+4x^2+C

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integral of 8x(x^2+1)^3

\int 8 x (x^{2} + 1)^{3} d x

x^{8} + 4 x^{6} + 6 x^{4} + 4 x^{2} + C

Solution steps

\int 8 x (x^{2} + 1)^{3} d x

Take the constant out:

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

= 8 \cdot \int x (x^{2} + 1)^{3} d x

Expand

x (x^{2} + 1)^{3} : x^{7} + 3 x^{5} + 3 x^{3} + x

= 8 \cdot \int x^{7} + 3 x^{5} + 3 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 x^{7} d x + \int 3 x^{5} d x + \int 3 x^{3} d x + \int x d x)

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

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

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

= 8 (\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2})

Simplify

8 (\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2}) : x^{8} + 4 x^{6} + 6 x^{4} + 4 x^{2}

= x^{8} + 4 x^{6} + 6 x^{4} + 4 x^{2}

Add a constant to the solution

= x^{8} + 4 x^{6} + 6 x^{4} + 4 x^{2} + C

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What is the integral of 8x(x^2+1)^3 ?
The integral of 8x(x^2+1)^3 is x^8+4x^6+6x^4+4x^2+C