What is the integral of 20x^3(5x^4-1)^3 ?

The integral of 20x^3(5x^4-1)^3 is 5/4 x^4(125x^{12}-100x^8+30x^4-4)+C

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integral of 20x^3(5x^4-1)^3

\int 20 x^{3} (5 x^{4} - 1)^{3} d x

\frac{5}{4} x^{4} (125 x^{12} - 100 x^{8} + 30 x^{4} - 4) + C

Solution steps

\int 20 x^{3} (5 x^{4} - 1)^{3} d x

Take the constant out:

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

= 20 \cdot \int x^{3} (5 x^{4} - 1)^{3} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}} ( 5 u ^{\frac{4}{3}} - 1 ) ^{3}}{3} d u

= 20 \cdot \int \frac{u ^{\frac{1}{3}} ( 5 u ^{\frac{4}{3}} - 1 ) ^{3}}{3} d u

Take the constant out:

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

= 20 \cdot \frac{1}{3} \cdot \int u^{\frac{1}{3}} (5 u^{\frac{4}{3}} - 1)^{3} d u

Expand

u^{\frac{1}{3}} (5 u^{\frac{4}{3}} - 1)^{3} : 125 u^{\frac{13}{3}} - 75 u^{3} + 15 u^{\frac{5}{3}} - u^{\frac{1}{3}}

= 20 \cdot \frac{1}{3} \cdot \int 125 u^{\frac{13}{3}} - 75 u^{3} + 15 u^{\frac{5}{3}} - 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

= 20 \cdot \frac{1}{3} (\int 125 u^{\frac{13}{3}} d u - \int 75 u^{3} d u + \int 15 u^{\frac{5}{3}} d u - \int u^{\frac{1}{3}} d u)

\int 125 u^{\frac{13}{3}} d u = \frac{375}{16} u^{\frac{16}{3}}

\int 75 u^{3} d u = \frac{75 u ^{4}}{4}

\int 15 u^{\frac{5}{3}} d u = \frac{45}{8} u^{\frac{8}{3}}

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

= 20 \cdot \frac{1}{3} (\frac{375}{16} u^{\frac{16}{3}} - \frac{75 u ^{4}}{4} + \frac{45}{8} u^{\frac{8}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = x^{3}

= 20 \cdot \frac{1}{3} (\frac{375}{16} (x^{3})^{\frac{16}{3}} - \frac{75 ( x ^{3} ) ^{4}}{4} + \frac{45}{8} (x^{3})^{\frac{8}{3}} - \frac{3}{4} (x^{3})^{\frac{4}{3}})

Simplify

20 \cdot \frac{1}{3} (\frac{375}{16} (x^{3})^{\frac{16}{3}} - \frac{75 ( x ^{3} ) ^{4}}{4} + \frac{45}{8} (x^{3})^{\frac{8}{3}} - \frac{3}{4} (x^{3})^{\frac{4}{3}}) : \frac{5}{4} x^{4} (125 x^{12} - 100 x^{8} + 30 x^{4} - 4)

= \frac{5}{4} x^{4} (125 x^{12} - 100 x^{8} + 30 x^{4} - 4)

Add a constant to the solution

= \frac{5}{4} x^{4} (125 x^{12} - 100 x^{8} + 30 x^{4} - 4) + C

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What is the integral of 20x^3(5x^4-1)^3 ?
The integral of 20x^3(5x^4-1)^3 is 5/4 x^4(125x^{12}-100x^8+30x^4-4)+C