x-5-x-3-1-1-3-dx-

Question Number 11149 by suci last updated on 14/Mar/17

\int x^{5} \sqrt[3]{x^{3} + 1} d x = \dots ? ? ?

Answered by ajfour last updated on 14/Mar/17

\frac{{(x^{3} + 1)}^{7 / 3}}{7} - \frac{{(x^{3} + 1)}^{4 / 3}}{4} + C

Commented by ajfour last updated on 14/Mar/17

l e t x^{3} + 1 = t

3 x^{2} d x = d t

s o, \int x^{5} {(x^{3} + 1)}^{1 / 3} d x

= \frac{1}{3} \int x^{3} {(x^{3} + 1)}^{1 / 3} 3 x^{2} d x

= \frac{1}{3} \int (t - 1) t^{1 / 3} d t = \frac{1}{3} \int (t^{4 / 3} - t^{1 / 3}) d t

= \frac{1}{3} (\frac{t^{7 / 3}}{7 / 3}) - \frac{1}{3} (\frac{t^{4 / 3}}{4 / 3}) + C

h e n c e t h e a n s w e r .

Answered by Mechas88 last updated on 17/Mar/17

u = {(x^{3} + 1)}^{1 / 3}

u^{3} = x^{3} + 1

x^{3} = u^{3} - 1

d u = \frac{1}{3} {(x^{3} + 1)}^{- 2 / 3} (3 x^{2}) d x = x^{2} {(x^{3} + 1)}^{- 2 / 3} d x

d x = x^{- 2} {(x^{3} + 1)}^{2 / 3} d u

\int x^{5} x^{- 2} u d u = \int (u^{3} - 1) u d u =

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

\frac{u^{5}}{5} - \frac{u^{2}}{2} + C =

\frac{\sqrt[3]{{(x^{3} + 1)}^{5}}}{5} - \frac{\sqrt[3]{{(x^{3} + 1)}^{2}}}{2} + C * * * R t a

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