How do you evaluate the integral #int x^2/(16+x^2)#?

Question

5625 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-20T01:45:48

I got:

#x - 4arctan(x/4) + C#

Using a trick:

#int (16 + x^2 - 16)/(16 + x^2)dx#

#= int dx - 16int 1/(16 + x^2)dx#

#= int dx - int 1/(1 + (x/4)^2)dx#

Now, for the second integral only, we would let #u = x/4# to find that #du = 1/4dx#. Thus, we multiply by #4# to maintain the integrand:

#= int dx - 4int 1/4 1/(1 + (x/4)^2)dx#

#= int dx - 4int 1/(1 + u^2)dx#

Now, if we recognize that #d/(dx)[arctanx] = 1/(1+x^2)#, then we have:

#=> int dx - 4arctanu#

#= color(blue)(x - 4arctan(x/4) + C)#