The integral of $\displaystyle x^3/\sqrt(1+x)$
How do you start this integral since it is an improper fraction. I tried squaring the top and bottom, but that didn't seem to work. What else should I try
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The integral of $\displaystyle x^3/\sqrt(1+x)$
How do you start this integral since it is an improper fraction. I tried squaring the top and bottom, but that didn't seem to work. What else should I try
A substitution. Let $\displaystyle u^2 = 1 + x$Quote:
Originally Posted by stevo970
Can you finish?
correction $\displaystyle x^3/\sqrt(1+x^2)$
That's even better! we won't have to cube a binomial.Quote:
Originally Posted by stevo970
Similar approach, let $\displaystyle u^2 = 1 + x^2$
so would you then have to figure out
du=$\displaystyle x/\sqrt(1+x^2)$dx or is that not the next step
That's not the next step...Quote:
Originally Posted by stevo970
$\displaystyle u^2 = 1 + x^2 \implies \boxed{x^2 = u^2 - 1}$
$\displaystyle \Rightarrow 2u~du = 2x~dx$
$\displaystyle \Rightarrow u~du = x~dx$
Now substitute and finish up
