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

The integral of 2x^3(x^4-4)^4 is 1/10 x^4(x^{16}-20x^{12}+160x^8-640x^4+1280)+C

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

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

\frac{1}{10} x^{4} (x^{16} - 20 x^{12} + 160 x^{8} - 640 x^{4} + 1280) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

u^{\frac{1}{3}} (u^{\frac{4}{3}} - 4)^{4} : u^{\frac{17}{3}} - 16 u^{\frac{13}{3}} + 96 u^{3} - 256 u^{\frac{5}{3}} + 256 u^{\frac{1}{3}}

= 2 \cdot \frac{1}{3} \cdot \int u^{\frac{17}{3}} - 16 u^{\frac{13}{3}} + 96 u^{3} - 256 u^{\frac{5}{3}} + 256 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

= 2 \cdot \frac{1}{3} (\int u^{\frac{17}{3}} d u - \int 16 u^{\frac{13}{3}} d u + \int 96 u^{3} d u - \int 256 u^{\frac{5}{3}} d u + \int 256 u^{\frac{1}{3}} d u)

\int u^{\frac{17}{3}} d u = \frac{3}{20} u^{\frac{20}{3}}

\int 16 u^{\frac{13}{3}} d u = 3 u^{\frac{16}{3}}

\int 96 u^{3} d u = 24 u^{4}

\int 256 u^{\frac{5}{3}} d u = 96 u^{\frac{8}{3}}

\int 256 u^{\frac{1}{3}} d u = 192 u^{\frac{4}{3}}

= 2 \cdot \frac{1}{3} (\frac{3}{20} u^{\frac{20}{3}} - 3 u^{\frac{16}{3}} + 24 u^{4} - 96 u^{\frac{8}{3}} + 192 u^{\frac{4}{3}})

Substitute back

u = x^{3}

= 2 \cdot \frac{1}{3} (\frac{3}{20} (x^{3})^{\frac{20}{3}} - 3 (x^{3})^{\frac{16}{3}} + 24 (x^{3})^{4} - 96 (x^{3})^{\frac{8}{3}} + 192 (x^{3})^{\frac{4}{3}})

Simplify

2 \cdot \frac{1}{3} (\frac{3}{20} (x^{3})^{\frac{20}{3}} - 3 (x^{3})^{\frac{16}{3}} + 24 (x^{3})^{4} - 96 (x^{3})^{\frac{8}{3}} + 192 (x^{3})^{\frac{4}{3}}) : \frac{1}{10} x^{4} (x^{16} - 20 x^{12} + 160 x^{8} - 640 x^{4} + 1280)

= \frac{1}{10} x^{4} (x^{16} - 20 x^{12} + 160 x^{8} - 640 x^{4} + 1280)

Add a constant to the solution

= \frac{1}{10} x^{4} (x^{16} - 20 x^{12} + 160 x^{8} - 640 x^{4} + 1280) + C

What is the integral of 2x^3(x^4-4)^4 ?
The integral of 2x^3(x^4-4)^4 is 1/10 x^4(x^{16}-20x^{12}+160x^8-640x^4+1280)+C