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

The integral of x^3(2x^4+4)^2 is 1/3 (x^{12}+6x^8+12x^4)+C

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

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

\frac{1}{3} (x^{12} + 6 x^{8} + 12 x^{4}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}} ( 2 u ^{\frac{4}{3}} + 4 ) ^{2}}{3} d u

= \int \frac{u ^{\frac{1}{3}} ( 2 u ^{\frac{4}{3}} + 4 ) ^{2}}{3} d u

Take the constant out:

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

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

Expand

u^{\frac{1}{3}} (2 u^{\frac{4}{3}} + 4)^{2} : 4 u^{3} + 16 u^{\frac{5}{3}} + 16 u^{\frac{1}{3}}

= \frac{1}{3} \cdot \int 4 u^{3} + 16 u^{\frac{5}{3}} + 16 u^{\frac{1}{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 4 u^{3} d u + \int 16 u^{\frac{5}{3}} d u + \int 16 u^{\frac{1}{3}} d u)

\int 4 u^{3} d u = u^{4}

\int 16 u^{\frac{5}{3}} d u = 6 u^{\frac{8}{3}}

\int 16 u^{\frac{1}{3}} d u = 12 u^{\frac{4}{3}}

= \frac{1}{3} (u^{4} + 6 u^{\frac{8}{3}} + 12 u^{\frac{4}{3}})

Substitute back

u = x^{3}

= \frac{1}{3} ((x^{3})^{4} + 6 (x^{3})^{\frac{8}{3}} + 12 (x^{3})^{\frac{4}{3}})

Simplify

\frac{1}{3} ((x^{3})^{4} + 6 (x^{3})^{\frac{8}{3}} + 12 (x^{3})^{\frac{4}{3}}) : \frac{1}{3} (x^{12} + 6 x^{8} + 12 x^{4})

= \frac{1}{3} (x^{12} + 6 x^{8} + 12 x^{4})

Add a constant to the solution

= \frac{1}{3} (x^{12} + 6 x^{8} + 12 x^{4}) + C

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What is the integral of x^3(2x^4+4)^2 ?
The integral of x^3(2x^4+4)^2 is 1/3 (x^{12}+6x^8+12x^4)+C