Question

Inverse Trig Integrals. Could someone show me how to do this one?

integral x*dx/(3+2x+2x^2)

Answer 1

id like to help but can you send me a photo of the original question?

Answer 2

\[
\int \frac{x\;dx}{3+2x+2x^2}
\]

Answer 3

thanks wio

ear, is this what you want to integrate

Answer 4

Yes, what wio wrote is exactly it! I can't break the denominator into u^2 + a^2

Answer 5

\[\int\limits \frac{ x~dx }{ 3+2 x+2 x^2 }=\frac{ 1 }{ 4 } \int\limits \frac{ 4x+2-2 }{ 3+2 x+2 x^2 }dx\]
\[=\frac{ 1 }{ 4 }\int\limits \frac{ 4x+2 }{ 3+2x+2x^2 }dx-\frac{ 1 }{ 2 }\int\limits \frac{ dx }{ 2\left( x^2+x+\frac{ 1 }{ 4 } -\frac{ 1 }{ 4 }\right)+3 }\]
\[=\frac{ 1 }{ 4 }\ln \left| 3+2x+2x^2 \right|-\frac{ 1 }{ 2 }\int\limits \frac{ dx }{ 2\left( x+\frac{ 1 }{ 2 } \right)^2-\frac{ 2 }{ 4 } +3}\]
i think now you can solve.

Answer 6

Thank you, surjithayer!

Answer 7

yw

Answer 8

nicely done!

Answer 9

what about the rest of it?

Answer 10

I'm getting back into finishing this problem now. I understand how you got to the part you showed me last night but I can't seem to figure out my substitution. I have attached the format it needs to be at the end. Thanks for any help!

Answer 11

correction you are correct it is +ve not -ve.
it is
\[\frac{ 1 }{ 4 } \int\limits \frac{ dx }{ \left( x+\frac{ 1 }{ 2 } \right)^2+\left( \frac{ \sqrt{5} }{ 2 } \right)^2 }\]
now you can write the answer straightway by the above formula

Answer 12

if you want substitution
put\[x+\frac{ 1 }{ 2 }=\frac{ \sqrt{5} }{ 2 }\tan \theta,dx=\frac{ \sqrt{5} }{ 2 }\sec ^2\theta~ d \theta \]

Answer 13

\[=\frac{ 1 }{ 4 }*\frac{ 1 }{ \frac{ \sqrt{5} }{ 2 } }\tan^{-1} \frac{ x+\frac{ 1 }{ 2 } }{ \frac{ \sqrt{5} }{ 2 } } +c\]
simplify it.

Answer 14

i hope now you got it.

Answer 15

Thanks again, Surjithayer!