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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x}

\frac{d}{d x} (5 e^{- x} + e^{3 x})

\frac{d}{d x} (ln (\frac{x}{1 + x ^{2}}))

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

y = 14 x, y = x^{2}

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