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If g(x)=∫dxx1/2+x1/6g(x) = \int \frac{dx}{x^{1/2} + x^{1/6}}, then g(1)−g(0)g(1) - g(0) is :

Solution

✅ Correct Option: 4

Let x=t6x=t^6, dx=6t5dtdx=6t^5 dt. Then x1/2=t3x^{1/2}=t^3, x1/6=tx^{1/6}=t, denominator =t(t2+1)= t(t^2+1). Integral =6∫t4t2+1dt=6∫(t2−1+1t2+1)dt=2t3−6t+6arctan⁡t= 6\int \frac{t^4}{t^2+1}dt = 6\int(t^2-1+\frac{1}{t^2+1})dt = 2t^3 - 6t + 6\arctan t. Evaluating from t=0t=0 to t=1t=1: 2−6+3π2=−4+3π22 - 6 + \frac{3\pi}{2} = -4 + \frac{3\pi}{2}.

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