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.
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
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
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
A + B = 0
C = 0
A = 2
Eh, not too bad after all. We thus have that
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
Let
=> 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"")
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
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
Let's try
=int(tan^2theta-1)/(tantheta(tan^2theta+1))(sec^2thetad theta)
Since
=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
=-lnabs(x/(x^2+1))+C=lnabs((x^2+1)/x)+C