How do you evaluate the integral #int 1/(xsqrt(2+x^2))#?

Question

Charles Anctil · Accepted Answer

#1/sqrt2lnabs(x/(sqrt(x^2+2)+sqrt2))+C#

#I=intdx/(xsqrt(2+x^2))#

Use the substitution #x=sqrt2tantheta#. This implies that #x^2=2tan^2theta# and #dx=sqrt2sec^2thetad theta#.

Plugging this in shows that:

#I=int(sqrt2sec^2thetad theta)/(sqrt2tanthetasqrt(2+2tan^2theta))#

Note that #sqrt(2+2tan^2theta)=sqrt2sqrt(1+tan^2theta)=sqrt2sectheta#. This comes from the identity #1+tan^2theta=sec^2theta#.

#I=int(sqrt2sec^2thetad theta)/(sqrt2tantheta(sqrt2sectheta))=1/sqrt2intsectheta/tanthetad theta#

Rewriting with #sintheta# and #costheta#:

#I=1/sqrt2int1/costhetacostheta/sinthetad theta=1/sqrt2intcscthetad theta=-1/sqrt2lnabs(csctheta+cottheta)#

From the original substitution we have #tantheta=x/sqrt2#. This correlates to a right triangle with the side opposite #theta# being #x# and the side adjacent to #theta# being #sqrt2#. The Pythagorean theorem gives the hypotenuse as #sqrt(x^2+2)#.

Thus, #csctheta# is the hypotenuse over the opposite side, or #csctheta=sqrt(x^2+2)/x# and #cottheta=1/tantheta=sqrt2/x#.

So:

#I=-1/sqrt2lnabs((sqrt(x^2+2)+sqrt2)/x)+C#

We can rewrite this using #Bln(A)=ln(A^B)#. In this case, we will use #-ln(A)=ln(A^-1)=ln(1/A)#.

#I=1/sqrt2lnabs(x/(sqrt(x^2+2)+sqrt2))+C#

Samantha Adkison · Answer

undefined To evaluate the integral ∫(1 / (x√(2 + x^2))) dx, you can use a trigonometric substitution. Let $ x = \sqrt{2} \tan(\theta) $. Then, $ dx = \sqrt{2} \sec^2(\theta) d\theta $. Substitute these into the integral and simplify it to make it easier to solve. Finally, integrate the resulting expression in terms of $ \theta $ and then substitute back in terms of $ x $ to obtain the final answer.