Another definite integral...

We can try expanding the denominator, using partial fractions to break it up, or do a trig-substitution. Both expanding the denominator and using partial fractions looks like it'll be messy, so let's try the trig-substitution.

$\bullet x = tan(\theta) \implies dx = sec^2(\theta) d\theta$

$\bullet \int_{x=0}^{\lim_{x \to \infty}} \frac{1}{(1+x^2)^3} dx = \int_{\theta=0}^{\lim_{\theta \to \frac{\pi}{2}}} \frac{sec^2(\theta)}{sec^6(\theta)} d\theta = \Large \int_{\theta=0}^{\lim_{\theta \to \frac{\pi}{2}}} cos^4(\theta) d\theta$

Now, by using the trig-identity $cos^2(\theta) = \frac{1}{2}(1+cos(2\theta))$ , we can just expand the whole thing out.

$\bullet cos^4(\theta) = (\frac{1}{2}(1+cos(2\theta)))^2 = \frac{1}{4}(1+2cos(2\theta)+cos^2(2\theta)) = \Large \frac{1}{4}(\frac{3}{2}+2cos(2\theta)+\frac{1}{2}cos(4\theta))$

So now we have the integral

$\bullet \frac{1}{4} \int_{\theta=0}^{\lim_{\theta \to \frac{\pi}{2}}} (\frac{3}{2}+2cos(2\theta)+\frac{1}{2}cos(4\theta)) d\theta$

$\bullet \frac{1}{4}(\frac{3}{2}\theta + sin(2\theta) + \frac{1}{8}sin(4\theta) |_0^{\frac{\pi}{2}} ) = \Large \boxed{\frac{3\pi}{16}}$

X=tan¥ Integral reduces to int lim 0- 1.78 Now using gamma functions calculating we get the answer