Is there a simple way to solve #int(x^2-1)/(x^3+x)dx#?

There is supposed to be a simple manipulation to this one that makes the answer really simple but I'm not sure what to do. I've tried everything I can think of including any trig subs that might have worked.

3 Answers
Truong-Son N.
Jun 6, 2017

Correct me if I'm wrong, but I don't think we could have done trig sub here since there aren't any square roots (or ways to obtain square root forms). We needed the form #sqrt(a^2 - x^2)#, #sqrt(a^2 + x^2)#, or #sqrt(x^2 - a^2)#, and we didn't have any form of that.

I got

#ln|(x^2 + 1)/x| + C#

There is somewhat of a trick I can think of off-hand, though it will lead to partial fractions... good practice if you want to do it this way.

#= int (x^2 + 1 - 2)/(x(x^2 + 1))dx#

#= int cancel(x^2 + 1)/(xcancel((x^2 + 1))) - 2/(x(x^2+1))dx#

#= int 1/x - 2/(x(x^2+1))dx#

But now it gets somewhat ugly as we get into partial fractions on the second integral, and the second factor is an irreducible quadratic (over the real numbers).

#int 1/x dx - ul(int 2/(x(x^2+1))dx) = ln |x| - ul(int A/(x) + (Bx + C)/(x^2 + 1)dx)#

For now, let's drop the integral signs and focus on the integrand. We obtain common denominators:

#(A(x^2 + 1) + (Bx + C)(x))/cancel(x(x^2 + 1)) = (2)/cancel(x(x^2 + 1))#

#Ax^2 + A + Bx^2 + Cx = 2#

#color(green)((A + B))x^2 + color(green)(C)x + color(green)(A) = color(green)(0)x^2 + color(green)(0)x + color(green)(2)#

We have matched the form #ax^2 + bx + c#, so we have the system of equations:

#A + B = 0#
#C = 0#
#A = 2#

Eh, not too bad after all. We thus have that #B = -2#. So:

#ln |x| - int A/(x) + (Bx + C)/(x^2 + 1)dx#

#= ln |x| - int 2/(x) - (2x)/(x^2 + 1)dx#

This leaves us with one integral left we don't immediately know. But, this looks like a #u# substitution.

Let #u = x^2 + 1# so that #du = 2xdx#. Then:

#=> ln |x| - 2 ln|x| + int 1/u du#

#= ln |x| - 2 ln|x| + ln|u| + C#

#= -ln|x| + ln|x^2 + 1| + C#

#= color(blue)(ln|(x^2 + 1)/x| + C)#

We can verify this is correct as well.

#d/(dx)[ln|x^2 + 1| - ln|x|]#

#= 1/(x^2 + 1) cdot 2x - 1/x#

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

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

#= (x^2 - 1)/(x^3 + x)# #color(blue)(sqrt"")#

John D.
Jun 9, 2017

Ah, I think I see what to do now that I know what the integral is.

Explanation:

The manipulation is simply multiplying the integral by #x^-2/x^-2#. This produces the following:

#int((x^2-1)(x^-2))/((x^3+x)(x^-2))dx#

#=int(1-x^-2)/(x+x^-1)dx#

And then I can make the substitution:

#u = x + x^-1#

#therefore du = (1 - x^-2)dx#

#therefore int(1-x^-2)/(x+x^-1)dx = int (du)/u = lnabs(u)+C#

#lnabs(x+1/x)+C#

Final Answer

mason m
Jun 9, 2017

#int(x^2-1)/(x^3+x)dx=int(x^2-1)/(x(x^2+1))dx#

Let's try #x=tantheta# so #dx=sec^2thetad theta#.

#=int(tan^2theta-1)/(tantheta(tan^2theta+1))(sec^2thetad theta)#

Since #tan^2theta+1=sec^2theta#:

#=int(tan^2theta-1)/tanthetad theta#

#=int(tantheta-cottheta)d theta#

#=int(sintheta/costheta-costheta/sintheta)d theta#

#-lnabscostheta-lnabssintheta+C#

#=-lnabs(sinthetacostheta)+C#

From #tantheta=x//1# we see that #sintheta=x/sqrt(x^2+1)# and #costheta=1/sqrt(x^2+1)#:

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

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