Without using substitution, I need to evaluate the following integral:

\[\int\limits_{}^{} \frac{ 1 }{ 1 + (3x)^{2} }dx \]

I know that the integral of 1/x is ln(x), so I feel like it would just be something simple like \[\ln(1+3x ^{2})\]

What's the derivative of \(\ln(1+3x^2)\)?

without using a substitution, you just have to recognize it

what has a derivative \(\frac{1}{1+x^2}\) ?

I believe this is an arctan integral. You need a 3 in the numerator so take out a 1/3.

I want to say that it's arctanx @satelite73

That's correct.

oh heck, sorry satellite, I didn't see your post trying to get the recognition of the derivative.

Mmmk, so \[\int\limits_{}^{} \frac{ 1 }{ 1 + (3x ^{2}) } \] \[\frac{ 1 }{ 3} \int\limits_{}^{} \frac{ 1 }{ 1 + x ^{2} }\] \[\frac{ 1 }{ 3 } \tan^{-1} x + c\]

Would that be right?

not quite the 3x has to remain in the final expression. The 1/3 comes from the fact that the derivative of 3x is 3 so we need a 3 in the numerator and thus need to take out a 1/3.

You arent supposed to use a u-sub, but you have to think that if it was to be used, u would be 3x. and thus du would be 3dx.

So where would I throw the 3x whilst writing the steps out? Take it out of the integral and then after I integrate just write arctan(3x) instead of arctan(x)?

It's more like this:\[\int\limits_{}^{}\frac{ 1 }{ 1+(3x)^2 }=\frac{ 1 }{ 3 }\tan^{-1}(3x)\] @alaskamath

Therefore, it would look like \[\frac{ 1 }{ 3 } \int\limits \frac{ du }{ 1+u ^{2} }=\frac{ 1 }{ 3 }\tan^{-1} u+c\] and then just replace u with 3x.

as genius suggests

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