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[Solved] Integration by substitution

Posted: Sun Feb 25, 2007 8:13 pm
by Ambush Commander
How would I integrate this function:

Code: Select all

x * sqrt(2x + 1)
using integration by substitution? If I set u = 2x + 1, I get du = 2dx, which doesn't cause the lone x to cancel out...

Edit Solved: the trick was getting x = (u - 1)/2 and substituting that into the new integrand with respect to u.