Integration of dy/(1+0.01y²)=dx

[asdf1]

In the problem, y`= 1+0.01y^2
the first i took was
dy/(1+0.01y^2)=dx
however, I'm stuck on the integration step...
can someone help?

 

[TD]

$$\int {\frac{{dy}}{{1 + 0.01y^2 }}} = \int {\frac{{dy}}{{1 + \left( {\frac{y}{{10}}} \right)^2 }}} = 10\int {\frac{{d\left( {\frac{y}{{10}}} \right)}}{{1 + \left( {\frac{y}{{10}}} \right)^2 }}}$$

The last one smells arctan-ish

 

[asdf1]

jeepers! thanks for reminding me about the arctan-ish thingy...
crud, totally forgot about that one~
:)

 

[TD]

No problem; but never forget the arctan hehe