## thomas5267 2 years ago 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.

1. Callisto

Completing square!?

2. Callisto

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

3. thomas5267

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

4. Callisto

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

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

6. thomas5267

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

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

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

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

10. thomas5267

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

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

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

13. thomas5267

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

14. Callisto

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

Did... I make it worse??

16. thomas5267

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

17. Callisto

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

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

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

20. thomas5267

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

21. Callisto

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

22. thomas5267

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

23. Callisto

Welcome :)