What is the integral of 2x(x^2+3)^2 ?

The integral of 2x(x^2+3)^2 is (x^6)/3+3x^4+9x^2+C

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integral of 2x(x^2+3)^2

\int 2 x (x^{2} + 3)^{2} d x

\frac{x ^{6}}{3} + 3 x^{4} + 9 x^{2} + C

Solution steps

\int 2 x (x^{2} + 3)^{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} + 3)^{2} d x

Expand

x (x^{2} + 3)^{2} : x^{5} + 6 x^{3} + 9 x

= 2 \cdot \int x^{5} + 6 x^{3} + 9 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 6 x^{3} d x + \int 9 x d x)

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 9 x d x = \frac{9 x ^{2}}{2}

= 2 (\frac{x ^{6}}{6} + \frac{3 x ^{4}}{2} + \frac{9 x ^{2}}{2})

Simplify

2 (\frac{x ^{6}}{6} + \frac{3 x ^{4}}{2} + \frac{9 x ^{2}}{2}) : \frac{x ^{6}}{3} + 3 x^{4} + 9 x^{2}

= \frac{x ^{6}}{3} + 3 x^{4} + 9 x^{2}

Add a constant to the solution

= \frac{x ^{6}}{3} + 3 x^{4} + 9 x^{2} + C

24 y^{^{''}} + 12 y^{^{'}} = 0, y (3) = - 3, y^{^{'}} (- 4) = 4

(2 x + 9)^{2}

\int \frac{tan ( x )}{1 - sin ^{2} ( x )} d x

f (x) = lo g_{e} (10 x)

\frac{d y}{d x} = sin^{2} (y)

What is the integral of 2x(x^2+3)^2 ?
The integral of 2x(x^2+3)^2 is (x^6)/3+3x^4+9x^2+C