# integration by substitution

• April 1st 2008, 05:29 PM
samdmansam
integration by substitution
int(cuberoot(x^3+1)x^5dx
• April 1st 2008, 05:32 PM
Jhevon
Quote:

Originally Posted by samdmansam
int(cuberoot(x^3+1)x^5dx

please clarify, you are missing some parenthesis. do you mean $\int \sqrt[3]{x^3 + 1} \cdot x^5~dx$ ?
• April 1st 2008, 05:37 PM
samdmansam
Int by substitution
Quote:

Originally Posted by Jhevon
please clarify, you are missing some parenthesis. do you mean $\int \sqrt[3]{x^3 + 1} \cdot x^5~dx$ ?

yes thats exactly what I meant, Thanks
• April 1st 2008, 05:39 PM
Jhevon
Quote:

Originally Posted by samdmansam
yes thats exactly what I meant, Thanks

ok, well integration by substitution will work nicely. make a substitution of $u = x^3 + 1$ (and note that $x^5 = x^3 \cdot x^2$)
• April 1st 2008, 05:51 PM
Krizalid
Make $z^3=x^3+1\implies z^2\,dz=x^2\,dx,$ it's a more direct way to solve the problem, you get rid of the cube root.
• April 1st 2008, 06:45 PM
samdmansam
could you please take it a little further
my text has very basic examples then skips to questions with cuberoots