Integration of 1/[ x^2*√(x^2+1)]?

1 Answer
Nov 17, 2017

#=-sqrt(x^2+1)/x+C#

Explanation:

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#