Integral of (1 + arctan(x/3))/(9 + x²) from 0 to 3

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theRukus
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Homework Statement


I missed one class on trigonometric identities in integrals, and I feel that one is needed here:

[itex]\int^3_0\frac{1+arctan(\frac{x}{3})}{9+x^2}dx[/itex]

Homework Equations


The Attempt at a Solution


Again, I'm unsure what to do. I think that it is a trig identity, but I could be wrong. I'll continue to try parts & substitutions..

Thanks!
 
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The derivative of acrtan(x) is 1/(x2+1), so the derivative of acrtan(x/3) is 3/(x2+9)
 
theRukus said:

Homework Statement


I missed one class on trigonometric identities in integrals, and I feel that one is needed here:

[itex]\int^3_0\frac{1+arctan(\frac{x}{3})}{9+x^2}dx[/itex]


Homework Equations





The Attempt at a Solution


Again, I'm unsure what to do. I think that it is a trig identity, but I could be wrong. I'll continue to try parts & substitutions..

Thanks!
Split the integral into two integrals. The first can be done directly and the second can be done with an ordinary substitution. Integration by parts is not the way to go.

Do you know this differentiation formula?
[tex]\frac{d}{dx} tan^{-1}(x)?[/tex]
 
Notice that [itex]\displaystyle \frac{d((f(x)^2)}{dx}=2f(x)f'(x)\,.[/itex]

That does that imply regarding [itex]\displaystyle \int\ f(x)f\,'(x)\,dx\,?[/itex]