Correct answer: (1) \(\frac{\pi}{8}\)\(\Big(1 - \frac{\sqrt 3}{2}\Big)\)

Explanation:

I = \(\int_0^1 \frac{\sqrt x}{(1+x)(1+3x)(3+x)}\)dx

Let x = t² ⇒ dx = 2t.dt

I = \(\int_0^1 \frac{t(2t)}{(t^2 + 1)(1+3t^2)(3+t^2)}\)dt

I = \(\int_0^1 \frac{(3t^2 + 1)-(t^2+1)}{(3t^2 + 1)(t^2 + 1)(3+t^2)}\)dt

I = \(\int_0^1 \frac{dt}{(t^2 + 1)(3+t^2)}\)- \(\int_0^1 \frac{dt}{(1+3t^2)(3+t^2)}\)

= \(\frac{1}{2}\)\(\int_0^1 \frac{(3+t^2)-(t^2+1)}{(t^2 + 1)(3+t^2)}\)dt + \(\frac{1}{8}\)\(\int_0^1 \frac{(1+3t^2)-3(3 + t^2)}{(1+ 3t^2)(3+t^2)}\)dt

= \(\frac{1}{2}\)\(\int_0^1 \frac{dt}{t^2 + 1}\)- \(\frac{1}{2}\)\(\int_0^1 \frac{dt}{t^2 + 3}\)+ \(\frac{1}{8}\)\(\int_0^1 \frac{dt}{t^2 + 3}\) - \(\frac{3}{8}\)\(\int_0^1 \frac{dt}{1+3t^2}\)

= \(\frac{1}{2}\)\(\int_0^1 \frac{dt}{t^2 + 1}\)- \(\frac{3}{8}\)\(\int_0^1 \frac{dt}{t^2 + 3}\) - \(\frac{3}{8}\)\(\int_0^1 \frac{dt}{1+3t^2}\)

= \(\frac{1}{2}(tan^{-1}(t))_0^1 - \frac{3}{8 \sqrt 3}\)\(\Big(tan^{-1}\Big(\frac{t}{\sqrt 3}\Big) \Big)_0^1\) - \(\frac{3}{8 \sqrt 3}\)\((tan^{-1}(\sqrt 3 t))_0^1\)

= \(\frac{1}{2}(\frac{\pi}{4})\)-\(\frac{\sqrt 3}{8}(\frac{\pi}{6})\) - \(\frac{\sqrt{3}}{8}(\frac{\pi}{3})\)

= \(\frac{\pi}{8}\) - \(\frac{\sqrt 3}{16} \pi\)

= \(\frac{\pi}{8}\)\(\Big(1 - \frac{\sqrt 3}{2}\Big)\)