1. calculus BC

i've been staring at these problems for at least an hour and still don't know where to start
can anyone give me some hints?

1.) integrate [(x^2)/(2+3x)^(1/2)] dx
2.) integrate [(x^2-25)^(1/2)/(x)] dx
i think this problem might be arcsin?? i don't know

these problems are either integration by parts (q*r-Sr*dq) or right triangles (trig)

thank you so much for any help!

2. After staring for an hour, you should take a break.

For the first, try $u^{2} = 2+3x$. It's not too bad. No Arcsine or anything past algebra.

For the second, perhaps $u^{2} = x^{2} - 25$ will lead somewhere useful.

3. yeah that's what i thought initially, but we are supposed to use integration by parts or right triangles

4. Originally Posted by holly123
i've been staring at these problems for at least an hour and still don't know where to start
can anyone give me some hints?

1.) integrate [(x^2)/(2+3x)^(1/2)] dx
2.) integrate [(x^2-25)^(1/2)/(x)] dx
i think this problem might be arcsin?? i don't know

these problems are either integration by parts (q*r-Sr*dq) or right triangles (trig)

thank you so much for any help!
(2) Try letting $x = 5 \sec \theta \Rightarrow \frac{1}{25}x^2-1=\frac{1}{25}25 \sec^2 \theta-1 = \tan^2 \theta$

Then $\int \frac{\sqrt{x^2-25}}{x}=\frac{\sqrt{25(\frac{1}{25}x^2-1)}}{x}=...$

5. Originally Posted by holly123
yeah that's what i thought initially, but we are supposed to use integration by parts or right triangles
Who say's you won't get around to that after the initial substitution?