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

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

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integral of 2x(x^2+1)

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{2} - 2, (2, 2)

\frac{d}{d x} (5 x arcsin (x))

x \to \infty lim (0, x)

n = 1 \sum \infty \frac{2}{3} (\frac{1}{3})^{n}

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