JEE Advanced 2018 — Paper 2 — Question 34

Evaluation of the Definite Integral

The given integral is:

$$I = \int_{0}^{1 / 2} \frac{1+\sqrt{3}}{\left((x+1)^{2}(1-x)^{6}\right)^{1 / 4}} \, dx$$

Simplify the Integrand We rewrite the expression inside the fourth root to isolate a power of $(1-x)$ that matches the derivative of a potential substitution:

$$(x+1)^{2}(1-x)^{6} = \frac{(x+1)^{2}}{(1-x)^{2}} \cdot (1-x)^{8} = \left(\frac{1+x}{1-x}\right)^{2} (1-x)^{8}$$

Applying the fourth root, we get:

$$\left[ \left(\frac{1+x}{1-x}\right)^{2} (1-x)^{8} \right]^{1/4} = \left(\frac{1+x}{1-x}\right)^{1/2} (1-x)^{2}$$

Substituting this back into the integral:

$$I = (1+\sqrt{3}) \int_{0}^{1 / 2} \frac{1}{(1-x)^{2} \sqrt{\frac{1+x}{1-x}}} \, dx$$

Substitution Let $u = \frac{1+x}{1-x}$. Then:

$$du = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^{2}} \, dx = \frac{2}{(1-x)^{2}} \, dx \implies \frac{dx}{(1-x)^{2}} = \frac{du}{2}$$

Changing the limits: As $x \to 0$, $u \to \frac{1+0}{1-0} = 1$ As $x \to 1/2$, $u \to \frac{1+1/2}{1-1/2} = 3$ Integration

$$I = \frac{1+\sqrt{3}}{2} \int_{1}^{3} u^{-1/2} \, du$$ $$I = \frac{1+\sqrt{3}}{2} \left[ 2\sqrt{u} \right]_{1}^{3}$$ $$I = (1+\sqrt{3})(\sqrt{3} - 1)$$

Final Result Using the difference of squares:

$$I = (\sqrt{3}+1)(\sqrt{3}-1) = (\sqrt{3})^{2} - (1)^{2} = 3 - 1 = 2$$

Final Answer: The value of the integral is $2$.