# Integration by Substitution problem

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?

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 $\sqrt{1-(\frac{x}{\sqrt {3}})^2}$

• 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 – Manny265 Nov 5 '13 at 1:15

HINT: Use the substitution $x=\sqrt{3}\cdot \sin(u)$.

• Does this help? – Chris K Nov 5 '13 at 0:38
• Here's another hint: this reduces to $5\cdot \sqrt{3}\cdot sin(u) + 3\; du$ – Chris K Nov 5 '13 at 0:40
• why such a substitution is there a logical explanation behind this? – Manny265 Nov 5 '13 at 1:16
• 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. – Chris K Nov 5 '13 at 1:27