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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

d er i v a t i v e sec (6 x)

n = 1 \sum \infty \frac{3}{( n + 1 ) !}

x \to 0 lim (\frac{3}{x} - \frac{3}{x ^{2} + x})

d er i v a t i v e 4

y^{^{'}} = \frac{3 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2}}{x ^{5} y}

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