What is the integral of 2x(x^2+4)^2 ?

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

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integral of 2x(x^2+4)^2

\int 2 x (x^{2} + 4)^{2} d x

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

Solution steps

\int 2 x (x^{2} + 4)^{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 (x^{2} + 4)^{2} d x

Expand

x (x^{2} + 4)^{2} : x^{5} + 8 x^{3} + 16 x

= 2 \cdot \int x^{5} + 8 x^{3} + 16 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^{5} d x + \int 8 x^{3} d x + \int 16 x d x)

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

\int 8 x^{3} d x = 2 x^{4}

\int 16 x d x = 8 x^{2}

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

Add a constant to the solution

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

t - (t - 5) (- (t - 5))

f (x) = 2 x^{2} - 7, at x = - 2

4 x + y^{2} = 13, x = 3 y

(ln (3 x))^{^{'}}

6 e^{- 5 t} + 5 t

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