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thomas5267
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 \(\int^b_a\frac{mt+c}{t^2-k^2}\). Not even a clue how to do it. Please help.
Consider the denominator: \[x^2+4x+5=(x^2 +4x+4-4)+5 = (x+2)^2 -4+5 =...\]
I don't think so. Completing the square will yield \((x+a)^2+b\), which is not what we want...
Hmm... \[\int^3_2\frac{3x}{x^2+4x+5} dx \]\[= \int^3_2\frac{3x}{x^2+4x+4-4+5} dx\]\[=\int^3_2\frac{3x}{(x+2)^2-4+5} dx\]\[=\int^3_2\frac{3x}{(x+2)^2+1} dx\]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
Ha! Maybe I'm not even going in the right direction!
OK, but then the problem tells us to integrate it. \[ \begin{align*} &\int^3_2\frac{3x}{(x+2)^2+1}\,dx\\ =&\int^5_3\frac{3t-6}{t^2+1}\,dt \end{align*} \] As far as I can see, this cannot be integrated, or is my brain not functioning properly now.
When x=2, t= 2+2 = 4... \[\int_4^5\frac{3t-6}{t^2+1}dt\]\[=\int_4^5\frac{3t}{t^2+1}-\frac{6}{t^2+1}dt\] 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.
OK, my brain is certainly not working properly. What is happening in here? I have never see something like this. \[ \int\frac{3t}{t^2+1}\,dt=\frac{3}{2}\int\frac{1}{t^2+1}\,d(t^2) \]
d/dt (t^2) = 2t, agree?
I still don't understand \(d(t^2)\) and what should I do with it. I would do this in the following way. \[ \begin{align*} &\int\frac{3t}{t^2+1}\,dt\\ =&3\int\frac{t}{t^2+1}\,dt\\ =&\frac{3}{2}\int\frac{1}{u}\,du\\ =&\frac{3}{2}\ln(|t^2+1|)+C \end{align*} \]
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.
But remember it's definite integral.. not indefinite..
Can you teach me how to do this in your way? It seems way faster then mine.
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
Did... I make it worse??
So I can just integrate the numerator? What should I do with \(\int\frac{1}{t^2+1}d(t^2)\) then?
Get the answer! ln|t^2 +1| +C! Just consider t^2 = u, it's simply \(\int \frac{1}{u+1}du\) , that is ln |u+1|+C = ln|t^2+1|+C
Is the following true? \[ \int^5_4\frac{3t}{t^2+1}dt = \frac{3}{2}\int^{25}_{16}\frac{1}{t^2+1}d(t^2) \]
No, you didn't change the variable, so you don't have to change the bound value.
What is the name of the procedure? I would like to find more information on the internet.
I'm sorry.. I don't know the name of it :(
Anyways, thank you! I have learnt a lot in here.