Question

integrate please.

Answer 1

please post you Q and we'll get right on it

Answer 2

\[\int\limits_{}^{} \frac{3x-5}{(x^2+4x+8)^\frac{3}{2}}\]dx

Answer 3

my answer is \[\frac{-5-x}{\sqrt{x^2+4x+8}} +C\]

Answer 4

Note that \(x^{2}+4x+8 = (x+2)^{2} + 2^{2}\) and hence try substituting \(x+2 = 2\tan{\theta}\)

and @kaiz122 , your answer does not give the right result (try differentiating your answer and see if you get back the original question)

Answer 5

i used trigonometric substitution here, i'll check back my solution,,

Answer 6

another way to start off (though perhaps less efficient)\[\int{3x-5\over x^2+4x+8}dx=\int{3x+6-11\over x^2+4x+8}dx\]\[=\int{3x+6\over x^2+4x+8}-{11\over x^2+4x+8}dx\]\[u=x^2+4x+8\implies du=2x+4dx\implies\frac32du=3x+6\]\[\frac32\int\frac{du}u-\int{11\over x^2+4x+8}dx\]complete the square on the second integral, etc.

Answer 7

@TuringTest , I believe you got the question wrong. It is \[\int \frac{3x-5}{(x^{2}+4x+8)^{3/2}}dx\] No doubt the first part would be as you have done (with minor changes), but then to integrate \[\int \frac{1}{(x^{2}+4x+8)^{3/2}}dx\]You again have to fall back on trigonometric substitution.

Answer 8

@FoolAroundMath you are right, I got hasty trying to avoid the trig sub :P
ignore my posts...

Answer 9

ok, thank you for your help, but i saw my error, i get 2 instead of 11, thank you

Answer 10

By the way, try wolfram first. The steps it gives are amazing and it shows the how and why.
http://www.wolframalpha.com/input/?i=integrate+%283x-5%29%2F%28x%5E%282%29%2B4x%2B8%29%5E%283%2F2%29

and then Show Steps