Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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.

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
Completing square!?
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...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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.
Welcome :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question