Evaluate the Integral (sqrt(1+x^2))/x dx Trigonometric Substitution.

[@adamding3873]

Use the replacement of tan(t)=x is not ideal in this case. Try x^2+1= t^2.
tdt = xdx
I = integral(t^2/x^2 dt) = integral(1 + 1/2(1/(t-1) - 1/(t+1))) dt = t + 1/2 ln|t-1| - 1/2 ln|t+1| = sqrt(x^2+1) + 1/2 ln|sqrt(x^2+1) -1| - 1/2 ln|sqrt(x^2+1) +1| + C
1/2 ln|sqrt(x^2+1) - 1| - 1/2 ln|sqrt(x^2+1) + 1| can be merged into
1/2 ln | (sqrt(x^2+1) - 1)/(sqrt(x^2+1) + 1) | = 1/2 ln | (sqrt(x^2+1) -1)^2 / (x^2 +1 -1 ) | = ln | (sqrt(x^2+1)-1 ) / x |
So the final result is the same.

[@aravjee8982]

thanks ma'am