What is the integral of (3x^3+1)^2*\sqrt[3]{x^7} ?

The integral of (3x^3+1)^2*\sqrt[3]{x^7} is 27/28 x^{28/3}+18/19 x^{19/3}+3/10 x^{10/3}+C

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integral of (3x^3+1)^2*\sqrt[3]{x^7}

\frac{27}{28} x^{\frac{28}{3}} + \frac{18}{19} x^{\frac{19}{3}} + \frac{3}{10} x^{\frac{10}{3}} + C

Solution steps

Apply u-substitution

: \int \frac{3 u ^{\frac{3}{7}} ( 3 u ^{\frac{9}{7}} + 1 ) ^{2}}{7} d u

= \int \frac{3 u ^{\frac{3}{7}} ( 3 u ^{\frac{9}{7}} + 1 ) ^{2}}{7} d u

Take the constant out:

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

= \frac{3}{7} \cdot \int u^{\frac{3}{7}} (3 u^{\frac{9}{7}} + 1)^{2} d u

Expand

u^{\frac{3}{7}} (3 u^{\frac{9}{7}} + 1)^{2} : 9 u^{3} + 6 u^{\frac{12}{7}} + u^{\frac{3}{7}}

= \frac{3}{7} \cdot \int 9 u^{3} + 6 u^{\frac{12}{7}} + u^{\frac{3}{7}} 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{3}{7} (\int 9 u^{3} d u + \int 6 u^{\frac{12}{7}} d u + \int u^{\frac{3}{7}} d u)

\int 9 u^{3} d u = \frac{9 u ^{4}}{4}

\int 6 u^{\frac{12}{7}} d u = \frac{42}{19} u^{\frac{19}{7}}

\int u^{\frac{3}{7}} d u = \frac{7}{10} u^{\frac{10}{7}}

= \frac{3}{7} (\frac{9 u ^{4}}{4} + \frac{42}{19} u^{\frac{19}{7}} + \frac{7}{10} u^{\frac{10}{7}})

Substitute back

Simplify

= \frac{27}{28} x^{\frac{28}{3}} + \frac{18}{19} x^{\frac{19}{3}} + \frac{3}{10} x^{\frac{10}{3}}

Add a constant to the solution

= \frac{27}{28} x^{\frac{28}{3}} + \frac{18}{19} x^{\frac{19}{3}} + \frac{3}{10} x^{\frac{10}{3}} + C

\int \frac{6 x ^{3} - 9 x + 7}{x ^{3} - x ^{2}} d x

\int x e x^{2} d x

\int \frac{x + 7}{x} d x

What is the integral of (3x^3+1)^2*\sqrt[3]{x^7} ?
The integral of (3x^3+1)^2*\sqrt[3]{x^7} is 27/28 x^{28/3}+18/19 x^{19/3}+3/10 x^{10/3}+C