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

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

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integral of 4(3+x^4)x

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

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

Solution steps

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

Take the constant out:

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

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

Expand

(3 + x^{4}) x : 3 x + x^{5}

= 4 \cdot \int 3 x + x^{5} 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

= 4 (\int 3 x d x + \int x^{5} d x)

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

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

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

Add a constant to the solution

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

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