int(cuberoot(x^3+1)x^5dx
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Originally Posted by samdmansam int(cuberoot(x^3+1)x^5dx please clarify, you are missing some parenthesis. do you mean $\displaystyle \int \sqrt[3]{x^3 + 1} \cdot x^5~dx$ ?
Originally Posted by Jhevon please clarify, you are missing some parenthesis. do you mean $\displaystyle \int \sqrt[3]{x^3 + 1} \cdot x^5~dx$ ? yes thats exactly what I meant, Thanks
Originally Posted by samdmansam yes thats exactly what I meant, Thanks ok, well integration by substitution will work nicely. make a substitution of $\displaystyle u = x^3 + 1$ (and note that $\displaystyle x^5 = x^3 \cdot x^2$)
Make $\displaystyle z^3=x^3+1\implies z^2\,dz=x^2\,dx,$ it's a more direct way to solve the problem, you get rid of the cube root.
my text has very basic examples then skips to questions with cuberoots
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