## 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 $$\int^b_a\frac{mt+c}{t^2-k^2}$$. Not even a clue how to do it. Please help. one year ago one year ago

1. Callisto Group Title

Completing square!?

2. Callisto Group Title

Consider the denominator: $x^2+4x+5=(x^2 +4x+4-4)+5 = (x+2)^2 -4+5 =...$

3. thomas5267 Group Title

I don't think so. Completing the square will yield $$(x+a)^2+b$$, which is not what we want...

4. Callisto Group Title

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

5. Callisto Group Title

Ha! Maybe I'm not even going in the right direction!

6. thomas5267 Group Title

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.

7. Callisto Group Title

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.

8. thomas5267 Group Title

OK, my brain is certainly not working properly. What is happening in here? I have never see something like this. $\int\frac{3t}{t^2+1}\,dt=\frac{3}{2}\int\frac{1}{t^2+1}\,d(t^2)$

9. Callisto Group Title

d/dt (t^2) = 2t, agree?

10. thomas5267 Group Title

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*}

11. Callisto Group Title

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.

12. Callisto Group Title

But remember it's definite integral.. not indefinite..

13. thomas5267 Group Title

Can you teach me how to do this in your way? It seems way faster then mine.

14. Callisto Group Title

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

15. Callisto Group Title

Did... I make it worse??

16. thomas5267 Group Title

So I can just integrate the numerator? What should I do with $$\int\frac{1}{t^2+1}d(t^2)$$ then?

17. Callisto Group Title

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

18. thomas5267 Group Title

Is the following true? $\int^5_4\frac{3t}{t^2+1}dt = \frac{3}{2}\int^{25}_{16}\frac{1}{t^2+1}d(t^2)$

19. Callisto Group Title

No, you didn't change the variable, so you don't have to change the bound value.

20. thomas5267 Group Title

What is the name of the procedure? I would like to find more information on the internet.

21. Callisto Group Title

I'm sorry.. I don't know the name of it :(

22. thomas5267 Group Title

Anyways, thank you! I have learnt a lot in here.

23. Callisto Group Title

Welcome :)