Finding the volume of the solid generated by revolving the given curve. The objective is to find the volume of the solid generated by revolving the curve $y=\dfrac{a^3}{a^2+x^2}$ about its asymptote.
Observing the given function yields that $y\ne0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about $x-axis$ is given as $$V=2\pi\int_0^{\infty}(f(x)^2-0)dx$$
Which gives: $2\pi\int_0^\infty\dfrac{a^6}{(a^2+x^2)^2}dx$
Now, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form $\dfrac{1}{x^4+2x^2+1}$ for $a=1$, which is transformed into $\dfrac{\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}=2[\dfrac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}-\dfrac{(1-\frac{1}{x^2})}{x^2+2+\frac{1}{x^2}}]$ which can be integrated, but this integral is tending to infinity. Can anyone help ? IS there a simpler way of doing this problem ?
 A: This seems like a trig substitution problem to me, albeit a not-so-straightforward one.  
Let $x = a \tan\theta$.  Then $dx = a \sec^2\theta\,d\theta$, and $a^2 + x^2 = a^2 \sec^2 \theta$.  Since $\theta = \tan^{-1}\left(\frac{x}{a}\right)$, $\theta = 0$ when $x=0$, and $\theta\to\frac{\pi}{2}$ as $x\to\infty$.  Therefore
\begin{align*}
    2\pi \int_0^\infty \frac{a^6}{(a^2 + x^2)^2}\,d\theta
    &= 2\pi \int_0^{\pi/2} \frac{a^6}{a^4\sec^4\theta}\,a \sec^2\theta \,d\theta\\
    &= 2\pi a^3 \int_0^{\pi/2} \cos^2\theta\,d\theta \\
    &=  \pi a^3 \int_0^{\pi/2}(1+\cos2\theta)\,d\theta \\
    &= \pi a^3 \left[\theta + \frac{1}{2}\sin2\theta\right]^{\pi/2}_0 \\
    &= \frac{\pi^2 a^3}{2}
\end{align*}
This agrees with the Wolfram Alpha answer.
I can't quite figure out how you intended to integrate $\int\frac{1 + \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx$, but it could be that you were trying to represent a convergent improper integral as the difference of two divergent improper integrals.  That leads to trouble, since $\infty - \infty$ is an indeterminate limit form.
A: Your integral is equal $\frac{\pi^2}{2}( \frac{1}{a^2})^{\frac{3}{2}}a^6$ according to wolfram alpha. Your algebra must be wrong somewhere. I recommend trying trigonometric substitutions.
A: Neither of your two integrals in the last step are divergent.  The first one is
$$
I_1 = \int_0^\infty\frac{1 + \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx = \int_0^\infty \frac{x^2 + 1}{x^4 + 2x^2 + 1} \, dx = \int_0^\infty \frac{dx}{x^2 + 1} = \left[\arctan x\right]_0^\infty = \frac{\pi}{2}.
$$
So that's nice and convergent.  The other one is a bit trickier, but we can find it with a quick jaunt into the complex plane: 
$$
\int \frac{1 - \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx = \int \frac{x^2 - 1}{x^4 + 2x^2 + 1} \, dx = \frac{1}{2} \left[ \int \frac{1}{(x+i)^2} + \int \frac{1}{(x-i)^2} \right] \\= -\frac{1}{2} \left[ \frac{1}{x + i} + \frac{1}{x - i} \right] = -\frac{1}{2} \frac{2x}{x^2 + 1} = - \frac{x}{x^2 + 1}
$$
and so
$$
\int_0^\infty \frac{1 - \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx = \left[ - \frac{x}{x^2 + 1} \right]_0^\infty = 0.
$$
