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

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

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

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

x^{2} + \frac{3 x ^{4}}{2} + x^{6} + \frac{x ^{8}}{4} + C

Solution steps

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

Take the constant out:

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

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

Expand

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

= 2 \cdot \int x + 3 x^{3} + 3 x^{5} + x^{7} 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

= 2 (\int x d x + \int 3 x^{3} d x + \int 3 x^{5} d x + \int x^{7} d x)

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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