We can use trigonometric substitution to solve this. Draw a right angle triangle. Name one acute angle #theta#. Label the side opposite to it #x# and the side adjacent to it #1#. From Pythagoras' formula which says #c^2=a^2+b^2# we know #c# is the hypotenuse, and #a# and #b# are the sides. We can write that equation for our triangle:
In our case, #a# is #x#, #b# is #1#, and #c# is the hypotenuse which we will calculate:
#c^2=x^2+1^2=x^2+1#
This gives us:
#c=sqrt(x^2+1)#
Then let's label the hypotenuse as that.
Now let's write the three basic trigonometric equations from this triangle:
#sintheta=x/sqrt(x^2+1)#
#costheta=1/(sqrt(x^2+1)#
#tantheta=x#
Now let's take derivative of both sides of the last equation:
#sec^2thetad(theta)=d(x)#
Our problem was:
#int1/(x^2sqrt(x^2+1))dx#
We can re-write it as:
#int(1/x^2)(1/sqrt(x^2+1))dx#
We now substitute trigonometric expressions for #x#:
#int1/tan^2theta*costhetasec^2thetad(theta)#
Now let's simplify. We know #tan=sin/cos# and #sec=1/cos#
#int(1/(sin^2theta/cos^2theta))*costheta*1/cos^2thetad(theta)#
#intcos^2theta/sin^2theta*costheta*1/cos^2thetad(theta)#
#intcos^3theta/(sin^2thetacos^2theta)d(theta)#
After we simplify we end up with:
#intcostheta/sin^2thetad(theta)#
Now we integrate by using the #u#-substitution method:
Let #u=sintheta#
#du=costhetad(theta)#
#intcostheta/sin^2thetad(theta)=int(costhetad(theta))/sin^2theta=int(du)/u^2=intu^-2du=#
#-u^-1+C=-1/u+C#
The #C# is there because we always add a constant to an indefinite integral. It is called indefinite when it does not have upper and lower bounds. Now we substitute back for #u#:
#=-1/sintheta+C#
Now we substitute back for #x#:
#=-1/(x/sqrt(x^2+1))+C#
#=-sqrt(x^2+1)/x+C#