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
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"")

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

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