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

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

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integral of (x^2+1)2x

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

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

Solution steps

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

Expand

(x^{2} + 1) x : 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

g (u) = 2 u + 3 u

- \frac{x}{1 - x ^{2}}

\int 1 + 36 x^{2} d x

\int (x \frac{e ^{- 2 x}}{2}) d x

y = sin (x) + tan (x)