Question

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.

Answer 1

Completing square!?

Answer 2

Consider the denominator:
\[x^2+4x+5=(x^2 +4x+4-4)+5 = (x+2)^2 -4+5 =...\]

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

I still don't understand \(d(t^2)\) and what should I do with it. I would do this in the following way.
\[
\begin{align*}
&\int\frac{3t}{t^2+1}\,dt\\
=&3\int\frac{t}{t^2+1}\,dt\\
=&\frac{3}{2}\int\frac{1}{u}\,du\\
=&\frac{3}{2}\ln(|t^2+1|)+C
\end{align*}
\]

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

Did... I make it worse??

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

Welcome :)