Question

Question #6c613

Answer 1

`arctan^2(2x)/4+C`

Explanation:

Note I'll be using the format `arctan(x)` instead of `"arctg"(x)`.

`intarctan(2x)/(1+4x^2)dx`

Note that since `d/dxarctan(x)=1/(1+x^2)`, we see that `d/dxarctan(2x)=1/(1+(2x)^2)d/dx(2x)=2/(1+4x^2)`.

So, if we try the substitution `u=arctan(2x)`, then `du=2/(1+4x^2)dx`. We currently have `1/(1+4x^2)dx` in the integrand, so we are just off by a factor of `2`.

`intarctan(2x)/(1+4x^2)dx=1/2intarctan(2x)2/(1+4x^2)dx=1/2intucolor(white).du`

Now use the rule `intu^ndu=u^(n+1)/(n+1)`:

`intarctan(2x)/(1+4x^2)dx=1/2u^2/2=u^2/4=arctan^2(2x)/4+C`