Question

The value of the integral \( \int_{0}^{1} \frac{\sqrt{x} dx}{(1+x)(1+3x)(3+x)} \) is :

• A. \(\frac{\pi}{8}\left(1 - \frac{\sqrt{3}}{2}\right)\)
• B. \(\frac{\pi}{8}\left(1 - \frac{\sqrt{3}}{6}\right)\)
• C. \(\frac{\pi}{4}\left(1 - \frac{\sqrt{3}}{6}\right)\)
• D. \(\frac{\pi}{4}\left(1 - \frac{\sqrt{3}}{2}\right)\)

Hint:

Integrals involving \(\sqrt{x}\) and rational functions of x often simplify with the substitution \(x=t^2\). This transforms the integrand into a standard rational function of t, which can then be tackled with partial fractions. Always double-check your final result against the options, as they might be written in a factored form.

Answer 1

The Correct Option is A

Step 1: Apply a substitution to simplify the integral.
Let \(x = t^2\). Then \(dx = 2t dt\). The limits of integration remain 0 to 1. The integral \(I\) becomes: \[ I = \int_{0}^{1} \frac{t \cdot (2t dt)}{(1+t^2)(1+3t^2)(3+t^2)} = \int_{0}^{1} \frac{2t^2 dt}{(1+t^2)(1+3t^2)(3+t^2)} \] Step 2: Use partial fraction decomposition.
Let \(y = t^2\). We decompose the integrand:
\[ \frac{2y}{(1+y)(1+3y)(3+y)} = \frac{A}{1+y} + \frac{B}{1+3y} + \frac{C}{3+y} \] Using the cover-up method to find the coefficients:
- For A (set y=-1): \(A = \frac{2(-1)}{(1+3(-1))(3-1)} = \frac{-2}{(-2)(2)} = \frac{1}{2}\)
- For B (set y=-1/3): \(B = \frac{2(-1/3)}{(1-1/3)(3-1/3)} = \frac{-2/3}{(2/3)(8/3)} = \frac{-1}{8/3} = -\frac{3}{8}\)
- For C (set y=-3): \(C = \frac{2(-3)}{(1-3)(1+3(-3))} = \frac{-6}{(-2)(-8)} = \frac{-6}{16} = -\frac{3}{8}\)
So, the integral in terms of t becomes:
\[ I = \int_{0}^{1} \left( \frac{1/2}{1+t^2} - \frac{3/8}{1+3t^2} - \frac{3/8}{3+t^2} \right) dt \] Step 3: Integrate each term.
We use the standard integral formulas \(\int \frac{du}{1+u^2} = \tan^{-1}(u)\) and \(\int \frac{du}{a^2+u^2} = \frac{1}{a}\tan^{-1}(u/a)\).
\[ I = \left[ \frac{1}{2}\tan^{-1}(t) - \frac{3}{8} \frac{\tan^{-1}(\sqrt{3}t)}{\sqrt{3}} - \frac{3}{8} \frac{\tan^{-1}(t/\sqrt{3})}{\sqrt{3}} \right]_0^1 \] The question seems to have a typo, and the intended denominator for the last term should likely lead to a cleaner result. However, assuming the coefficients are as calculated, let's re-examine the partial fraction for the intended answer. It is likely there is a combination of terms that simplifies better. The complexity suggests a possible error in the question's denominator. However, performing the integration as derived leads to: \[ I = \frac{1}{2}[\tan^{-1}(t)]_0^1 - \frac{\sqrt{3}}{8}[\tan^{-1}(\sqrt{3}t)]_0^1 - \frac{\sqrt{3}}{8}[\tan^{-1}(t/\sqrt{3})]_0^1 \] Evaluating at the limits t=1 and t=0: \[ I = \left( \frac{1}{2}(\frac{\pi}{4}) \right) - \left( \frac{\sqrt{3}}{8}(\frac{\pi}{3}) \right) - \left( \frac{\sqrt{3}}{8}(\frac{\pi}{6}) \right) \] \[ I = \frac{\pi}{8} - \frac{\sqrt{3}\pi}{24} - \frac{\sqrt{3}\pi}{48} = \frac{6\pi - 2\sqrt{3}\pi - \sqrt{3}\pi}{48} = \frac{6\pi - 3\sqrt{3}\pi}{48} = \frac{\pi(2 - \sqrt{3})}{16} \] Let's check option (A): \[ \frac{\pi}{8}\left(1 - \frac{\sqrt{3}}{2}\right) = \frac{\pi}{8}\left(\frac{2 - \sqrt{3}}{2}\right) = \frac{\pi(2 - \sqrt{3})}{16} \] The calculated result matches option (A).