`int sqrt(1 + x^2) / x dx` Evaluate the integral

`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