The value of the integral $I=\int\limits_0^a \frac{x^4}{\left(a^2+x^2\right)^4} d x$, is

$\frac{1}{16 a^3}\left(\frac{\pi}{4}-\frac{1}{3}\right)$

Putting $x=a \tan \theta$ in the given integral, we get

$x=a \tan \theta⇒dx=a\sec^2θdθ$

$I=\int\limits_0^a\frac{x^4dx}{(a^2+x^2)^4}⇒I=\int\limits_0^{\pi / 4}\frac{1}{a^3}\sin 4θ\cos^2θdθ$

$I=\int\limits_0^{\pi / 4}\frac{1}{a^3×8}(1-\cos 2θ)^2(\cos 2θ+1)dθ$

$I=\frac{1}{8a^3}\int\limits_0^{\pi / 4}(\sin^22θ)(1-\cos 2θ)dθ=\frac{1}{8a^3}\int\limits_0^{\pi / 4}\frac{(1-\cos 4θ)(1-\cos 2θ)dθ}{2}$

$I=\frac{1}{16a^3}\int\limits_0^{\pi / 4}(1-\cos 2θ-\cos 4θ+\cos 2θ×\cos 4θ)dθ$

$I=\frac{1}{32 a^3} \int\limits_0^{\pi / 4}(2-\cos 2 \theta-2 \cos 4 \theta+\cos 6 \theta) d \theta$

$\Rightarrow \ I=\frac{1}{32 a^3}\left[2 \theta-\frac{\sin 2 \theta}{2}-\frac{\sin 4 \theta}{2}+\frac{\sin 6 \theta}{6}\right]_0^{\pi / 4}=\frac{1}{16 a^3}\left(\frac{\pi}{4}-\frac{1}{3}\right)$