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Integrate ((cube root of x) - 1) / square root of x) with respect to x. Do I need to use substitution and what do I substitute?
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\[\int{\sqrt[3]x-1\over\sqrt x}dx\]if this is correct, rewrite this with fractional exponents and divide out the terms you can then integrate straightforward
yes, it's correct. I will try and if I fail, I shall return for help
I'll be around :)
\[6\div5 \times \sqrt[6]{x ^{5}} + 2\sqrt{x}\] is this the answer, do you maybe know?
sure, just don't forget the +C *always* include the +C for indefinite integrals
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oh yeah, thanks a lot :)
np :)
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