# integral { 1/(sqrt(x) * (1+x)) dx} for x out of R+using the substitution rule

*print*Print*list*Cite

Expert Answers

beckden | Certified Educator

Substitute `u=sqrt(x)` , `du=1/(2sqrt(x))dx`

This gives us `2du=1/sqrt(x) dx`

Our integral becomes

`int 1/(sqrt(x))*1/(x+1) dx = 2 int 1/(u^2+1) du`

Substituting `tan v = u` , `sec^2v dv = du` we get

`2 int 1/(u^2+1) du = 2 int 1/(tan^2v+1) sec^2v dv`

Since `tan^2v+1 = sec^2v` this integral becomes

`2 int dv = 2v + C` substituting u back in we get

`2v + C = 2 arctan u + C` and substituting x back in we get the final answers

`int (dx)/(sqrt(x)(x+1))=2 arctan(sqrt(x)) + C`