Is it possible in some cases that using the ILATE rule does not yield an explicit antiderivative but making another choice does yields one? If so, please give examples.
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What's the ILATE rule? – Hans Lundmark Sep 26 '14 at 12:51
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We learnt it as LIATE haha..Logs, inverse trig, algebra, trig, exponentials. – lagrange103 Sep 26 '14 at 12:53
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Yes, it's easy for the rule to fail if the proposed derivative is not integrable. For example in the integral
$$\int x^3 e^{x^2} \mathrm{d}x$$
the rule would propose $u=x^3$ and $dv=e^{x^2}$. The latter cannot be integrated and you are therefore stuck.
To solve the above integral use $u=x^2$ and $dv=x e^{x^2}$ instead. Then you get $du=2x$ and $v=\frac{1}{2}e^{x^2}$, leading to
$$\int x^3 e^{x^2} \mathrm{d}x=\\ \frac{x^2}{2}e^{x^2} -\int 2x \frac{1}{2}e^{x^2} \mathrm{d}x = \\ \frac{x^2}{2}e^{x^2} - \frac{1}{2}e^{x^2}=\\ \frac{1}{2}e^{x^2}(x^2-1)$$
There are likely other types of examples, but I can't think of any at the moment.
StefanS
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Another example is $\int\frac{xe^x}{(x+1)^2} dx$.
According to this rule, you would let $u=\frac{x}{(x+1)^2}$ and $dv=e^x dx$,
which would give $\frac{xe^x}{(x+1)^2}-\int\frac{(1-x)e^x}{(x+1)^3}dx$.
Instead, you want to let $u=xe^x$ and $dv=\frac{1}{(x+1)^2}dx$,
which gives $-\frac{xe^x}{x+1}-\int -e^x dx$.
user84413
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