∫1/(x2+4)2dx

Question

what if the integration for this term?

Answer 1

Let x=2tan(y), dx=2sec²(y)dy. x²+4=4(tan²(y)+1)=4sec²(y).

1/(x²+4)²=1/(16sec⁴(y)), so the integrand becomes 2sec²(y)dy/(16sec⁴(y))=

dy/(8sec²(y))=(cos²(y)/8)dy. cos(2y)=2cos²(y)-1, so cos²y=(cos(2y)+1)/2.

We now have ∫(cos(2y)+1)dy/16, which integrates to (1/16)(sin(2y)/2+y)+C, where C is integration constant.

tan(y)=x/2, so sin(y)=x/√(x²+4) and cos(y)=2/√(x²+4), sin(2y)/2=sin(y)cos(y)=2x/(x²+4), y=arctan(x/2).

Therefore the integral is (1/16)(2x/(x²+4)+arctan(x/2)).