Solving an Integral Involving ∫(4x)/(x2+9) dx

[jdawg]

Homework Statement

∫(4x)/(x²+9) dx

Homework Equations

The Attempt at a Solution

Originally I tried to solve with u substitution:

u=x²+9
du=2x dx
1/2du=dx
∫2/u
=2lnu+C
=2ln(x²+9)+C

But shouldn't arctan be somewhere in the answer?

 

[jdawg]

Oops, I made a typo.
1/2xdu=dx

 

[SteamKing]

What's the derivative of an arctan function?

In any event, if you differentiate your answer, you should obtain the original integrand.

 

[jdawg]

Is it (1/1+x²)?

 

[Curious3141]

Is it (1/1+x²)?

Yes, assuming you meant $\frac{1}{1+x^2}$, because your parentheses are placed wrongly. But $\frac{x}{1+x^2}$ is a completely different expression from $\frac{1}{1+x^2}$. In this case, you have something more like the former. Hence no arctan in the integral.

 

[jdawg]

But I don't understand why u substitution doesn't work?

 

[Curious3141]

But I don't understand why u substitution doesn't work?

Who said it doesn't work? Your integral in the first post is correct.

 

[jdawg]

Who said it doesn't work? Your integral in the first post is correct.

Really?? When I went to a tutoring center at my university the tutor told me that I needed to use arctan? Or was she just saying that its another method of doing it?

 

[Chestermiller]

Really?? When I went to a tutoring center at my university the tutor told me that I needed to use arctan? Or was she just saying that its another method of doing it?

No. She was saying it because she was wrong.

 

[jdawg]

No. She was saying it because she was wrong.

Hahaha! Thanks for your help everybody :)

 

[gopher_p]

To be fair, it's possible that by "use arctan" the tutor meant to use a trig sub $\tan\theta=x$, which would be the same as $u=\arctan x$. This would work, though it'd most definitely be a more involved approach than the simple u-sub suggested in the thread.