Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

without using a substitution, you just have to recognize it

Answer 5

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

Answer 6

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

Answer 7

I want to say that it's arctanx @satelite73

Answer 8

That's correct.

Answer 9

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

Answer 10

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

Answer 11

Would that be right?

Answer 12

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.

Answer 13

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.

Answer 14

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)?

Answer 15

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

Answer 16

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.

Answer 17

as genius suggests