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I really need help with this trig substitution please. integral of (x^3)/sqrt(9(x^2)+49)

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You have to do two different substitutions... First, set u=x², so du=2xdx. The integral becomes:\[\int\limits_{}^{}\frac{ udu }{ \sqrt{9u+49 }}\]Next, set p = 9u+49, so dp=9du, giving:\[\frac{ 1 }{ 81 } \int\limits_{}^{}\frac{ p-49 }{ \sqrt{p} }dp\]Now you can split up the fraction:\[\frac{ 1 }{ 81 }\int\limits_{}^{}\left( p^{\frac{1}{2}}-49p^{-\frac{1}{2}} \right)dp\]This is not difficult anymore, I guess...
Oops, forgot to compensate with ½ for the 2x, so replace 1/81 by 1/162, to get even!
If that second step is a little hard to follow, then here's extra explanation: p=9u+49, so u= (p-49)/9 = 1/9 *(p-49). Also: dp=9du, so du = 1/9 * dp. That accounts for the 1/81. Extra factor 1/2 I forgot, gives you 1/162.

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@MajikDUSTY: I think it's your turn now ;)
perfect way and new to me. thanks for that
Question: do we have to be back to u and then to x to get the final answer or just stop at p, since we substitute and respect to u and then to p dp ? just a little bit confuse
No, you have to go back to x. First go from p to u, then from u to x, so in the end you'll have a nice (ahem) formula with x...
got it. I'm not asker. I read it and recognize that the way you solve the problem is really new to me. Just confirm the stuff. Thanks a lot
You're right. Glad to be of help.

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