What is the integral of 6x(x^2+1)2 ?

The integral of 6x(x^2+1)2 is 12((x^4)/4+(x^2)/2)+C

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integral of 6x(x^2+1)2

\int 6 x (x^{2} + 1) 2 d x

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

Solution steps

\int 6 x (x^{2} + 1) \cdot 2 d x

Take the constant out:

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

= 6 \cdot 2 \cdot \int x (x^{2} + 1) d x

Simplify

= 12 \cdot \int x (x^{2} + 1) d x

Expand

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

= 12 \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

= 12 (\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}

= 12 (\frac{x ^{4}}{4} + \frac{x ^{2}}{2})

Add a constant to the solution

= 12 (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

x \to 9 - lim (\frac{1}{( x - 9 ) ^{2}})

\frac{d}{d x} (3 x^{2} y - 2 y x^{2} + 4 x - 5 y)

\int \frac{1}{x 36 - x ^{2}} d x

x \to - π lim (cot (x))

\frac{1}{( s ^{2} ( s + 1 ) )}

What is the integral of 6x(x^2+1)2 ?
The integral of 6x(x^2+1)2 is 12((x^4)/4+(x^2)/2)+C