Q.Integrate the function: $\displaystyle \int \frac{1}{x^{1/2}+x^{1/3}}\,dx$
[Hint: put $x=t^{6}$]

Concept: U Substitution — the hint $x = t^6$ eliminates fractional exponents by making the denominator a polynomial in $t$.

Step 1: Substitute $x = t^6$, so $dx = 6t^5\,dt$. Then

$x^{1/2} = t^3$ and $x^{1/3} = t^2$.

Step 2: The integral becomes

Step 3: Perform polynomial division: $\frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1}$.

Thus …

Why this substitution works

When you see fractional powers like $x^{1/2}$ and $x^{1/3}$, the exponents have denominators 2 and 3. The least common multiple of 2 and 3 is 6. So if we set $x = t^6$, then:

• $x^{1/2} = (t^6)^{1/2} = t^3$
• $x^{1/3} = (t^6)^{1/3} = t^2$

Both become simple integer powers of $t$. The hint in the problem is exactly this — it’s the cleanest way to handle mixed fractional exponents.

1. Perform the substitution

   Let $x = t^6$. Then $dx = 6t^5\,dt$. The integral becomes:

2. Simplify the integrand

   Factor the denominator: $t^3 + t^2 = t^2(t + 1)$. So:

The $t^2$ cancels, leaving a much simpler rational function.

3. Perform polynomial division

   The numerator $t^3$ has a higher degree than the denominator $t+1$, so we divide:

Let’s verify: $(t+1)(t^2 - t + 1) = t^3 - t^2 + t + t^2 - t + 1 = t^3 + 1$. So indeed $t^3 = (t+1)(t^2 - t + 1) - 1$, giving the result above.

Therefore:

4. Integrate term by term

Simplify the coefficients:

5. Substitute back to $x$

   Since $t = x^{1/6}$, we have:

   • $t^3 = (x^{1/6})^3 = x^{1/2} = \sqrt{x}$ …

Method: LCM substitution $x=t^{k}$ to clear fractional exponents

Use this whenever the integrand mixes roots like $x^{1/2}$ and $x^{1/3}$: substitute $x=t^{k}$ with $k=\operatorname{lcm}$ of the denominators to turn everything into a rational function of $t$.

Steps

Step 1: Choose the exponent.

Take $k=\operatorname{lcm}(m,n)$ of the fractional-power denominators (here $\operatorname{lcm}(2,3)=6$), so every root becomes an integer power of $t$. Remember $dx=k\,t^{k-1}\,dt$.

Step 2: Simplify to a rational function. …

Common Mistakes

Mistake 1: Forgetting $dx=6t^5\,dt$.

Why it's wrong: with $x=t^6$ the differential is not $dt$; omitting the $6t^5$ changes the whole integrand. Correct approach: always transform $dx$ under a substitution.

Mistake 2: Not choosing the LCM exponent.

Why it's wrong: using $x=t^2$ or $x=t^3$ leaves one root fractional; only $x=t^{\operatorname{lcm}(2,3)}=t^6$ clears both. Correct approach: take the LCM of the denominators. …