Hi! Can someone please walk me through on how to answer these integrals?

1. Attachment 24718

2. Attachment 24719

3. Attachment 24720

Thanks! :-)

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- Sep 6th 2012, 06:53 PMHi888How to solve Integrals?
Hi! Can someone please walk me through on how to answer these integrals?

1. Attachment 24718

2. Attachment 24719

3. Attachment 24720

Thanks! :-) - Sep 6th 2012, 07:13 PMSworDRe: How to solve Integrals?
1. Make the substitution t = (u-3)^2 - 8

2. Try integration by parts, so that inside the integral sign, you differentiate ln(x) and integrate sqrt(x), the product of these two will be a power function.

3. Make the substitution s = u-3, multiply out the square, and then you will have a sum of power functions. - Sep 6th 2012, 07:20 PMProve ItRe: How to solve Integrals?
Here's an easier substitution in the first...

$\displaystyle \displaystyle \begin{align*} \int{\frac{1}{3 + \sqrt{t + 8}}\,dt} &= \int{\frac{2\sqrt{t + 8}}{2\sqrt{t + 8}\left( 3 + \sqrt{t+ 8}\right)} \,dt} \\ &= \int{\frac{2u}{3 + u}\,du} \textrm{ after making the substitution }u = \sqrt{t + 8} \implies du = \frac{1}{2\sqrt{t + 8}}\,dt \\ &= \int{\frac{2(v - 3)}{v}\,dv} \textrm{ after making the substitution } v = 3 + u \implies dv = du \\ &= \int{\frac{2v - 6}{v}\,dv} \\ &= \int{2 - \frac{6}{v}\,dv} \end{align*}$

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