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I was given an integration problems sheet...with answers too but how a certain answer is to be obtained is obviuosly not stated.
Using integration by substitution integrate the following:
$$ \int \dfrac{5x +3}{\sqrt{3-x^2}} \, dx$$


And the answer at the back is:
$$ -5 \sqrt{3-x^2} +\arcsin \left( \frac{x}{\sqrt{3}}\right) +C$$
Any idea how i go about the substitution?

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First note that

$$ \int \frac{5x +3 }{\sqrt{3-x^2}}dx= \int \frac{5x }{\sqrt{3-x^2}}dx +\int \frac{ 3 }{\sqrt{3-x^2}}dx$$

For the first integral you can use the substitution $u=3-x^2$ and for the second, express the denominator as $ \sqrt{1-(\frac{x}{\sqrt {3}})^2} $

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  • $\begingroup$ used $u^2 = 3-x^2$ and that worked fine. Thanx alot,when u do things u think are complicated u tend to forget that there are some other simpler ways to view things from ie breaking it into fractions $\endgroup$ – Manny265 Nov 5 '13 at 1:15
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HINT: Use the substitution $x=\sqrt{3}\cdot \sin(u)$.

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  • $\begingroup$ Does this help? $\endgroup$ – Chris K Nov 5 '13 at 0:38
  • $\begingroup$ Here's another hint: this reduces to $5\cdot \sqrt{3}\cdot sin(u) + 3\; du$ $\endgroup$ – Chris K Nov 5 '13 at 0:40
  • $\begingroup$ why such a substitution is there a logical explanation behind this? $\endgroup$ – Manny265 Nov 5 '13 at 1:16
  • $\begingroup$ Well... you have to evaluate $x'(u)$, since we have to convert the differential form from $dx$ to $du$ in the numerator. So, it is kind of nice when we get the derivative in the denominator. $\endgroup$ – Chris K Nov 5 '13 at 1:27

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