What is the integral of 2(3x^4+1)^2 ?

The integral of 2(3x^4+1)^2 is 2(x^9+(6x^5)/5+x)+C

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integral of 2(3x^4+1)^2

\int 2 (3 x^{4} + 1)^{2} d x

2 (x^{9} + \frac{6 x ^{5}}{5} + x) + C

Solution steps

\int 2 (3 x^{4} + 1)^{2} d x

Take the constant out:

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

= 2 \cdot \int (3 x^{4} + 1)^{2} d x

Expand

(3 x^{4} + 1)^{2} : 9 x^{8} + 6 x^{4} + 1

= 2 \cdot \int 9 x^{8} + 6 x^{4} + 1 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 9 x^{8} d x + \int 6 x^{4} d x + \int 1 d x)

\int 9 x^{8} d x = x^{9}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 1 d x = x

= 2 (x^{9} + \frac{6 x ^{5}}{5} + x)

Add a constant to the solution

= 2 (x^{9} + \frac{6 x ^{5}}{5} + x) + C

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What is the integral of 2(3x^4+1)^2 ?
The integral of 2(3x^4+1)^2 is 2(x^9+(6x^5)/5+x)+C