∫\frac{x+x^{2/3}+x^{1/6}}{x(1+x^{1/3})}dx=px^{2/3}+q\tan^{-1}(x^{1/6})+c, \text{then :-}
Step-by-Step Solution
Key Concept: Substitute x = t^6, dx = 6t^5 dt. The integral becomes \int(t^6 + t^4 + t) / (t^6(1 + t^2)) * 6t^5 dt = 6 \int(t^5 + t^3 + 1) / (t(1 + t^2)) dt = 6 \int(t^4(t) + t^2(t) + 1) / (t(1 + t^2)) dt. Simplify to 6 \int(t^4 + t^2 + 1/t) / (1 + t^2) dt = 6 \int(t^2(t^2+1) + 1/t) / (1 + t^2) dt = 6 \int(t^2 + 1/(t(1+t^2))) dt = 6 \int(t^2 + 1/t - t/(1+t^2)) dt. Integrating gives 6(t^3/3 + ln|t| - 1/2 ln(1+t^2)) + c = 2t^3 + 6ln|t| - 3ln(1+t^2) + c. Re-evaluating the original integral: \int(x + x^(2/3) + x^(1/6)) / (x(1 + x^(1/3))) dx. Let x^(1/6) = t, x = t^6, dx = 6t^5 dt. Integral = \int(t^6 + t^4 + t) / (t^6(1 + t^2)) * 6t^5 dt = 6 \int(t^6 + t^4 + t) / (t(1 + t^2)) dt = 6 \int(t^5 + t^3 + 1) / (1 + t^2) dt = 6 \int(t^3(t^2+1) + 1) / (1 + t^2) dt = 6 \int(t^3 + 1/(1+t^2)) dt = 6(t^4/4 + tan^(-1)t) + c = 3/2 t^4 + 6 tan^(-1)t + c. Substituting t = x^(1/6), we get 3/2 x^(2/3) + 6 tan^(-1)(x^(1/6)) + c. Thus p = 3/2, q = 6. Since 1.5 < 6, p < q.
Let $x=t^6$, then $dx=6t^5dt$. The integral becomes $\int \frac{t^6+t^4+t}{t^6(1+t^2)} 6t^5 dt = 6 \int \frac{t^6+t^4+t}{t(1+t^2)} dt = 6 \int \frac{t^5+t^3+1}{1+t^2} dt = 6 \int \left( t^3 + \frac{1}{1+t^2} \right) dt = 6 \left( \frac{t^4}{4} + \tan^{-1}t \right) + c = \frac{3}{2} t^4 + 6 \tan^{-1}t + c = \frac{3}{2} x^{2/3} + 6 \tan^{-1}(x^{1/6}) + c$. Comparing with $px^{2/3} + q \tan^{-1}(x^{1/6}) + c$, we get $p = 3/2$ and $q = 6$. Since $1.5 < 6$, $p < q$.
Correct Answer: A