What is the integral of 8x^3(2x^4+4)^4 ?

The integral of 8x^3(2x^4+4)^4 is 32/5 x^4(x^{16}+10x^{12}+40x^8+80x^4+80)+C

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integral of 8x^3(2x^4+4)^4

\int 8 x^{3} (2 x^{4} + 4)^{4} d x

\frac{32}{5} x^{4} (x^{16} + 10 x^{12} + 40 x^{8} + 80 x^{4} + 80) + C

Solution steps

\int 8 x^{3} (2 x^{4} + 4)^{4} d x

Take the constant out:

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

= 8 \cdot \int x^{3} (2 x^{4} + 4)^{4} d x

Apply u-substitution

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

= 8 \cdot \int \frac{u ^{\frac{1}{3}} ( 2 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

= 8 \cdot \frac{1}{3} \cdot \int u^{\frac{1}{3}} (2 u^{\frac{4}{3}} + 4)^{4} d u

Expand

u^{\frac{1}{3}} (2 u^{\frac{4}{3}} + 4)^{4} : 16 u^{\frac{17}{3}} + 128 u^{\frac{13}{3}} + 384 u^{3} + 512 u^{\frac{5}{3}} + 256 u^{\frac{1}{3}}

= 8 \cdot \frac{1}{3} \cdot \int 16 u^{\frac{17}{3}} + 128 u^{\frac{13}{3}} + 384 u^{3} + 512 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

= 8 \cdot \frac{1}{3} (\int 16 u^{\frac{17}{3}} d u + \int 128 u^{\frac{13}{3}} d u + \int 384 u^{3} d u + \int 512 u^{\frac{5}{3}} d u + \int 256 u^{\frac{1}{3}} d u)

\int 16 u^{\frac{17}{3}} d u = \frac{12}{5} u^{\frac{20}{3}}

\int 128 u^{\frac{13}{3}} d u = 24 u^{\frac{16}{3}}

\int 384 u^{3} d u = 96 u^{4}

\int 512 u^{\frac{5}{3}} d u = 192 u^{\frac{8}{3}}

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

= 8 \cdot \frac{1}{3} (\frac{12}{5} u^{\frac{20}{3}} + 24 u^{\frac{16}{3}} + 96 u^{4} + 192 u^{\frac{8}{3}} + 192 u^{\frac{4}{3}})

Substitute back

u = x^{3}

= 8 \cdot \frac{1}{3} (\frac{12}{5} (x^{3})^{\frac{20}{3}} + 24 (x^{3})^{\frac{16}{3}} + 96 (x^{3})^{4} + 192 (x^{3})^{\frac{8}{3}} + 192 (x^{3})^{\frac{4}{3}})

Simplify

8 \cdot \frac{1}{3} (\frac{12}{5} (x^{3})^{\frac{20}{3}} + 24 (x^{3})^{\frac{16}{3}} + 96 (x^{3})^{4} + 192 (x^{3})^{\frac{8}{3}} + 192 (x^{3})^{\frac{4}{3}}) : \frac{32}{5} x^{4} (x^{16} + 10 x^{12} + 40 x^{8} + 80 x^{4} + 80)

= \frac{32}{5} x^{4} (x^{16} + 10 x^{12} + 40 x^{8} + 80 x^{4} + 80)

Add a constant to the solution

= \frac{32}{5} x^{4} (x^{16} + 10 x^{12} + 40 x^{8} + 80 x^{4} + 80) + C

1 + (\frac{2}{5})^{\circ} x (\frac{2}{3})^{\circ} - 25

g (x) = [ln (5 x - 2)]^{4}

\frac{1}{12 x ^{3}}

\int_{0}^{\infty} \frac{4}{x ^{2} + 4} d x

\frac{d}{d x} (tan (e^{x}))

What is the integral of 8x^3(2x^4+4)^4 ?
The integral of 8x^3(2x^4+4)^4 is 32/5 x^4(x^{16}+10x^{12}+40x^8+80x^4+80)+C