Question

$\int\frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=Ax^{\frac{1}{2}}+Bx^{\frac{1}{3}}+Cx^{\frac{1}{6}}+D~log(x^{\frac{1}{6}}+1)+k$, then values of A, B, C and D are}

• A. 2, -3, 6, -6
• B. 2, 3, -6, 6
• C. 2, -3, -6, 6
• D. -2, -3, 6, 6

Hint:

For integrals with $x^{1/n}$ and $x^{1/m}$, substitute $x = t^{LCM(n, m)}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The integral involves fractional powers of $x$. We use a substitution to rationalise the integrand.

Step 2: Key Formula or Approach:
The LCM of the denominators of the powers $(2, 3)$ is $6$. So, let $x = t^6$.

Step 3: Detailed Explanation:
Let $x = t^6 \Rightarrow dx = 6t^5 dt$.
$x^{1/2} = t^3$ and $x^{1/3} = t^2$.
The integral becomes: \[ I = \int \frac{6t^5 dt}{t^3 + t^2} = \int \frac{6t^5}{t^2(t + 1)} dt = \int \frac{6t^3}{t + 1} dt \] Using long division or synthetic substitution: $t^3 = (t^3 + 1) - 1 = (t+1)(t^2 - t + 1) - 1$. \[ I = 6 \int (\frac{(t+1)(t^2 - t + 1)}{t+1} - \frac{1}{t+1}) dt \] \[ I = 6 \int (t^2 - t + 1 - \frac{1}{t+1}) dt \] \[ I = 6 (\frac{t^3}{3} - \frac{t^2}{2} + t - \log(t+1)) + k \] \[ I = 2t^3 - 3t^2 + 6t - 6 \log(t+1) + k \] Substituting $t = x^{1/6}$: \[ I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6 \log(x^{1/6} + 1) + k \] Comparing with the given form: $A=2, B=-3, C=6, D=-6$.

Step 4: Final Answer:
The values are $2, -3, 6, -6$.