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

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

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

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

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

Solution steps

\int 2 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

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

Expand

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

= 2 \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

= 2 (\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}

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

Simplify

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

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

Add a constant to the solution

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

\int sec^{3} (2 x) tan (2 x) d x

x^{2} (1 - x)^{2}

\int x^{2} e^{- 2 x^{2}} d x

y = 1 - x^{2}, y = 3 - 3 x

\frac{\partial}{\partial x} (2 x y^{4} - 2)

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