Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 19

intsqrt(1+x^2)/xdx
Let's evaluate using the trigonometric substitution,
Let x=tan(theta)
dx=sec^2(theta)d theta
=intsqrt(1+tan^2(theta))/(tan(theta))*sec^2(theta) d theta
Now use the identity: 1+tan^2(x)=sec^2(x)
=intsqrt(sec^2(theta))/tan(theta)*sec^2(theta)d theta
=intsec(theta)/tan(theta)*sec^2(theta)d theta
=int(sec(theta)(1+tan^2(theta)))/tan(theta) d theta
=int((sec(theta))/tan(theta)+(sec(theta)tan^2(theta))/tan(theta))d theta
=intsec(theta)/tan(theta)d theta+intsec(theta)tan(theta)d theta
=int(1/cos(theta))*(1/(sin(theta)/cos(theta)))d theta+intsec(theta)tan(theta)d theta
=int1/sin(theta) d theta+intsec(theta)tan(theta)d theta
=intcsc(theta)d theta+intsec(theta)tan(theta)d theta
Now using the standard integrals,
intcsc(x)dx=ln|csc(x)-cot(x)
intsec(x)tan(x)dx=sec(x)
=ln|csc(theta)-cot(theta)|+sec(theta)
Now substitute back x=tan(theta)
=>cot(theta)=1/tan(theta)=1/x
1+tan^2(theta)=sec^2(theta)
=>1+x^2=sec^2(theta)
=>sec(theta)=sqrt(1+x^2)
1+cot^2(theta)=csc^2(theta)
=>1+(1/x)^2=csc^2(theta)
csc(theta)=sqrt(1+x^2)/x
:.intsqrt(1+x^2)/xdx=ln|sqrt(1+x^2)/x-1/x|+sqrt(1+x^2)+C ,C is a constant