Prove the equality $$B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx$$ for $p > 0, q > 0$
I think the first step is
$$B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx = \int \limits_0^1 \frac{x^{p-1}}{(1+x)^{p+q}}dx + \int \limits_0^1 \frac{x^{q-1}}{(1+x)^{p+q}}dx$$
Let $I_1 = \displaystyle \int_{0}^{1} \frac{x^{p-1}}{(1+x)^{p+q}} \mathrm{d}x$. We now substitute $x=\tan^2 \theta$ and thus, we get, $$I_1 = \int_{0}^{\pi/4} \frac{\tan^{2p-2} \theta}{\sec^{2p+2q} \theta} 2\tan \theta \sec^2 \theta \,\,\mathrm{d}\theta$$ $$=2\int_{0}^{\pi/4} \tan^{2p-1}\theta \sec^{2-2p-2q} \theta \,\,\mathrm{d}\theta$$ $$I_1 = 2\int_{0}^{\pi/4} \sin^{2p-1} \theta \cos^{2q-1}\theta\,\, \mathrm{d}\theta$$
Similarly, $$I_2 = \int_{0}^{1} \frac{x^{q+1}}{(1+x)^{p+q}}\mathrm{d}x = 2\int_{0}^{\pi/4} \cos^{2p-1} \theta \sin^{2q-1}\theta\,\, \mathrm{d}\theta$$
Now, using the fact that $\mathrm {B}(p,q) = 2\int_{0}^{\pi/2} \sin^{2p-1}\theta \cos^{2q-1}\theta\,\,\mathrm{d}\theta$, $$I_1 + I_2 \implies ?$$
Hope you can take it from here.
Note that
$$\begin {align} 2\int_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx &=\int_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx+\int_1^\infty \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx\\&=\int_0^\infty \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx\\&=\int_0^\infty \frac{x^{p-1} }{(1+x)^{p+q}}dx+\int_0^\infty \frac{x^{q-1}}{(1+x)^{p+q}}dx \\&=B (p,q)+B(q,p)\\&=2 B(p,q) \end {align}$$