# Improper Integral

This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

$$I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}$$

which, after carrying out the 1.2, becomes

$$\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}$$

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

$$\frac{1}{x^{p}}$$

I started off by trying a direct comparison test and compared it to $$\frac{2x}{x^{2.4}}$$ which can simplify to

$$\frac{2}{x^{.4}}$$

So, since the above function has $$p=.4<1$$, it diverges, by the Direct Comparison Test, $$I$$ diverges.

Last edited:

Mute
Homework Helper
Be careful! $(a + b)^p \neq (a^{p} + b^p)$.

Furthermore, the function $2x/x^{2.4}$ is actually slightly larger than your original integrand, so the fact that the integral over this diverges doesn't actually tell you anything about the convergence/divergence of your original integrand.

The fact that there is a $2x$ in the integrand suggests you should make the substitution $u = x^2 + 1$. Then your integral becomes

$$I = \int^{\infty}_{1}\frac{du}{u^{1.2}} = \int^{\infty}_{1}\frac{du}{u^{6/5}}$$

This is an integral you can do, and you should be able to tell if it diverges or converges.

Last edited:
HallsofIvy
Homework Helper
This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

$$I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}$$

which, after carrying out the 1.2, becomes

$$\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}$$
No, it doesn't! In general, (x+ y)n is NOT xn+ yn!

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

$$\frac{1}{x^{p}}$$

I started off by trying a direct comparison test and compared it to $$\frac{2x}{x^{2.4}}$$ which can simplify to

$$\frac{2}{x^{.4}}$$

So, since the above function has $$p=.4<1$$, it diverges, by the Direct Comparison Test, $$I$$ diverges.
Let u= x2+ 1, so that du= 2x dx, in your original integral.