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thomas5267
Group Title
Rewrite \(\int^3_2\frac{3x}{x^2+4x+5}\,dx\) into the form of \(\int^b_a\frac{mt+c}{t^2+k^2}\) or \(\int^b_a\frac{mt+c}{t^2k^2}\).
Not even a clue how to do it. Please help.
 one year ago
 one year ago
thomas5267 Group Title
Rewrite \(\int^3_2\frac{3x}{x^2+4x+5}\,dx\) into the form of \(\int^b_a\frac{mt+c}{t^2+k^2}\) or \(\int^b_a\frac{mt+c}{t^2k^2}\). Not even a clue how to do it. Please help.
 one year ago
 one year ago

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Callisto Group TitleBest ResponseYou've already chosen the best response.3
Completing square!?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Consider the denominator: \[x^2+4x+5=(x^2 +4x+44)+5 = (x+2)^2 4+5 =...\]
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
I don't think so. Completing the square will yield \((x+a)^2+b\), which is not what we want...
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Hmm... \[\int^3_2\frac{3x}{x^2+4x+5} dx \]\[= \int^3_2\frac{3x}{x^2+4x+44+5} dx\]\[=\int^3_2\frac{3x}{(x+2)^24+5} dx\]\[=\int^3_2\frac{3x}{(x+2)^2+1} dx\]So, in this case, t= x+2, k=1 For the numerator, we want mt+c, and we have 3x. m must be 3. 3(x+2) +c = 3x c = 6
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Ha! Maybe I'm not even going in the right direction!
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
OK, but then the problem tells us to integrate it. \[ \begin{align*} &\int^3_2\frac{3x}{(x+2)^2+1}\,dx\\ =&\int^5_3\frac{3t6}{t^2+1}\,dt \end{align*} \] As far as I can see, this cannot be integrated, or is my brain not functioning properly now.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
When x=2, t= 2+2 = 4... \[\int_4^5\frac{3t6}{t^2+1}dt\]\[=\int_4^5\frac{3t}{t^2+1}\frac{6}{t^2+1}dt\] For \(\int\frac{3t}{t^2+1}dt\) \[\int\frac{3t}{t^2+1}dt = \frac{3}{2}\int\frac{1}{t^2+1}d(t^2) \]Shouldn't be difficult For\(\int\frac{6}{t^2+1}dt\), use trigo sub.
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
OK, my brain is certainly not working properly. What is happening in here? I have never see something like this. \[ \int\frac{3t}{t^2+1}\,dt=\frac{3}{2}\int\frac{1}{t^2+1}\,d(t^2) \]
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
d/dt (t^2) = 2t, agree?
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
I still don't understand \(d(t^2)\) and what should I do with it. I would do this in the following way. \[ \begin{align*} &\int\frac{3t}{t^2+1}\,dt\\ =&3\int\frac{t}{t^2+1}\,dt\\ =&\frac{3}{2}\int\frac{1}{u}\,du\\ =&\frac{3}{2}\ln(t^2+1)+C \end{align*} \]
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Basically, it's just substitution.. u = t^2 +1 du = 2t dt t dt = du/2 But you keep it as t, anyway, forget it! And you get it right.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
But remember it's definite integral.. not indefinite..
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
Can you teach me how to do this in your way? It seems way faster then mine.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Hmm... I just did integration for the numerator and put it after ''d'' \[\int\frac{3t}{t^2+1}dt\]If you integrate the numerator, you'll get \(\frac{t^2}{2}\)So, \[\int\frac{3t}{t^2+1}dt = 3\int\frac{t}{t^2+1}dt =3 \int\frac{1}{t^2+1}d(\frac{t^2}{2})=\frac{3}{2} \int\frac{1}{t^2+1}d(t^2) \] Another way to think about it is that, d/dt (t^2 /2) gives you t, the numerator
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Did... I make it worse??
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
So I can just integrate the numerator? What should I do with \(\int\frac{1}{t^2+1}d(t^2)\) then?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Get the answer! lnt^2 +1 +C! Just consider t^2 = u, it's simply \(\int \frac{1}{u+1}du\) , that is ln u+1+C = lnt^2+1+C
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
Is the following true? \[ \int^5_4\frac{3t}{t^2+1}dt = \frac{3}{2}\int^{25}_{16}\frac{1}{t^2+1}d(t^2) \]
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
No, you didn't change the variable, so you don't have to change the bound value.
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
What is the name of the procedure? I would like to find more information on the internet.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
I'm sorry.. I don't know the name of it :(
 one year ago

thomas5267 Group TitleBest ResponseYou've already chosen the best response.0
Anyways, thank you! I have learnt a lot in here.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Welcome :)
 one year ago
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