Evaluate the integrals [closed]

I have a few integrals here that I need help on, plz give me some hints or suggestions on how to solve them. Thanks so much!

1. $$\int \frac{7\ln x}{x\sqrt{2 + (\ln x)^2}}\ dx$$
2. $$\int \sqrt{\frac{9+x}{9-x}}\ dx$$
3. $$\int 4\sqrt{4 + e^x}\ dx$$
4. $$\int \frac{dx}{x\sqrt{8x + 49}}$$
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What have you done/tried/thought? This looks like homework so please tag it as such (don't worry, people will still try to help you out). It though is important to show some self effort. –  DonAntonio Oct 7 '12 at 4:16
You can try change of variables first for many integrals. –  Patrick Li Oct 7 '12 at 4:18
I think it would be much better if you would post one integral, indicate what you can do with it and where you get stuck, digest the answers you get, see if you can apply them to the others; if not, ask one more, rinse, repeat. –  Gerry Myerson Oct 7 '12 at 4:28
First thing I'd like to say (repeat): One integral per post plzzz~ –  FrenzY DT. Oct 7 '12 at 4:42
And then, try to integrate by substitution. –  FrenzY DT. Oct 7 '12 at 4:43
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closed as not constructive by Andres Caicedo, Norbert, Noah Snyder, tomasz, Davide GiraudoOct 10 '12 at 13:41

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4 Answers

Hint on the first one: try $u=\sqrt{2+(\log x)^2}$. Then $u^2=2+(\log x)^2$, $2u\,du=2(\log x)/x$, etc.

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I tried u=lnx and then applied trig substitution for the rest of the equation –  Jaden Q Oct 7 '12 at 5:46
$u=\log x$ works, but if you then use trig substitution, you're working too hard. –  Gerry Myerson Oct 7 '12 at 11:48
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HINT : The first one is straightforward if you use $\ln x = u$.

That is $$\int \frac{7\ln x}{x\sqrt{2 + (\ln x)^2}}\ dx=\int \frac{7u}{\sqrt{2 + u^2}}\ du=7\int (\sqrt{2 + u^2})'\ du=7 \sqrt{2 + u^2}=7 \sqrt{2 +(\ln x)^2}+C$$

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thanks! i appreciate your help! –  Jaden Q Oct 7 '12 at 6:51
@Jaden Q: you're welcome. –  Chris's sis Oct 7 '12 at 6:52
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1. $\frac{7}{2}\int \frac{d[2+(\ln x)^2]}{\sqrt{2+(\ln x)^2}}=\frac{7}{2}\int\frac{du}{\sqrt{u}}=...$

2. Substitute $x=9\cos2\theta$

3. Substitute $4+e^x=y^2$

4. $=\frac{8}{3}\int ue^{-u} du=...$

5. Substitute $\sqrt{8x+49}=y$

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First of all I apologize that my mathJax is not rendering, and my formula may fail.

I'd say by attempting $u$-substitution, the problem could be solved.

== Why $u$-substitute? ==

Then, why $u$-substitute? To change a strange-looking integrand to something you might recognize. Even if obscured at the beginning, try to substitute different expressions (either schematically or randomly). You may stumble upon the right answer.

Take this little exercise:

$$\int \frac{\cos x}{1+\sin x \,\mathrm d x}$$

There is a fraction. The denominator is the sum of $1$ and a function $\sin x$.

If you are experienced enough to notice that the derivative of $1+\sin x$ is $\cos x$, then you may simply let

$$u=1+\sin x$$ with $$\,\mathrm d u = \cos x \,\mathrm d x$$ so that

$$\int \frac{\cos x}{1+\sin x \,\mathrm d x} = \int\frac{du}{u}=\ln\left|u\right|+C$$

Well, you say where's the x? Using careful manipulation we chose to hide x by letting $u=1+\sin x$ there in the beginning. Now, it's high time that we substituted the $u$ into the R.H.S. of the last equation and obtain

$$\ln\left|1+\sin x\right| + C.$$

(example from book: A First Course in Mathematical Modelling, F. R. Giodano)

== Try different substitutions ==

The third integral of OP is not to easy at first. OP would try $u=e^x$, ($\,\mathrm d x = u^{-1} \,\mathrm d u$) but it doesn't make things easier. Try to put more things into your substitution, try a few more minutes please.

Spoiler alert!


Try to simplify the square root.

Spoiler alert!


Try $u=4+e^x$

Spoiler alert! Finish the first few spoilers and proceed.


Alternatively, try to differentiate $u^{3/2} \ln u$, and see what terms you need to add.

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