Question

\[ \int \frac{1}{x^{1/2}+x^{1/3}}\,dx \] is equal to:

• A. \( \sqrt{x}-\sqrt[3]{x}+\sqrt[6]{x}-\log|x^{1/6}+1|+C \)
• B. \( 2\sqrt{x}-3\sqrt[3]{x}+6\sqrt[6]{x}-6\log|x^{1/6}+1|+C \)
• C. \( 2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\log|x^{1/6}+1|+C \)
• D. \( \sqrt{x}+\sqrt[3]{x}+\sqrt[6]{x}+\log|x^{1/6}+1|+C \)

Hint:

For integrals containing several roots like: \[ x^{1/2},\ x^{1/3},\ x^{1/4} \] always substitute: \[ x=t^{\text{LCM of denominators}} \] This converts the integral into a rational algebraic form which becomes much easier to integrate.

Answer 1

The Correct Option is B

Concept: Whenever fractional powers such as: \[ x^{1/2},\quad x^{1/3} \] appear together, we use substitution: \[ x=t^n \] where \(n\) is the LCM of denominators. This converts irrational powers into ordinary polynomial expressions.

Step 1: Choosing a suitable substitution.
The denominators are: \[ 2,\quad 3 \] Their LCM is: \[ 6 \] Let: \[ x=t^6 \] Differentiate: \[ dx=6t^5dt \] Also: \[ x^{1/2}=t^3 \] and \[ x^{1/3}=t^2 \] Substitute into the integral: \[ I= \int \frac{6t^5}{t^3+t^2}\,dt \] Factorize denominator: \[ I= 6\int \frac{t^5}{t^2(t+1)}\,dt \] \[ I= 6\int \frac{t^3}{t+1}\,dt \]

Step 2: Performing polynomial division.
Divide \(t^3\) by \(t+1\): \[ \frac{t^3}{t+1} = t^2-t+1-\frac1{t+1} \] Thus: \[ I= 6\int \left( t^2-t+1-\frac1{t+1} \right)dt \]

Step 3: Integrating term by term.
\[ I= 6\left[ \frac{t^3}{3} -\frac{t^2}{2} +t -\log|t+1| \right]+C \] Multiply by \(6\): \[ I= 2t^3-3t^2+6t-6\log|t+1|+C \]

Step 4: Substituting back in terms of \(x\).
Since: \[ t=x^{1/6} \] we get: \[ t^3=x^{1/2}=\sqrt{x} \] \[ t^2=x^{1/3}=\sqrt[3]{x} \] Therefore: \[ I= 2\sqrt{x} -3\sqrt[3]{x} +6\sqrt[6]{x} -6\log|x^{1/6}+1| +C \] Hence, \[ \boxed{ 2\sqrt{x} -3\sqrt[3]{x} +6\sqrt[6]{x} -6\log|x^{1/6}+1| +C } \]