Q.Evaluate: $\int \dfrac{x^{1/2}}{1+x^{3/4}}\,dx$ (Hint: Put $x=z^4$)

Concept: Partial Fraction Decomposition — the substitution $x = z^4$ eliminates fractional powers, turning the integrand into a rational function.

Step 1: Let $x = z^4$, so $dx = 4z^3\,dz$. Then $x^{1/2} = z^2$ and $x^{3/4} = z^3$.

Step 2: The integral becomes

Step 3: Perform polynomial division: $\frac{z^5}{1+z^3} = z^2 - \frac{z^2}{1+z^3}$.

Thus …

The given integral has a mix of fractional powers: $x^{1/2}$ and $x^{3/4}$. The common denominator of the exponents (2 and 4) is 4, so the substitution $x = z^4$ will clear all radicals and turn the integrand into a rational function of $z$. That’s the core idea — rational functions are much easier to integrate, especially with partial fractions.

Let’s work through it step by step.

1. Substitute $x = z^4$. Then $dx = 4z^3\,dz$. Also, $x^{1/2} = (z^4)^{1/2} = z^2$, and $x^{3/4} = (z^4)^{3/4} = z^3$. The integral becomes:

2. Simplify the rational function. The degree of the numerator (5) is greater than the degree of the denominator (3), so we must divide. Perform polynomial division: $z^5 \div (z^3 + 1)$.

Check: $(z^3+1)z^2 = z^5 + z^2$, subtract from $z^5$ gives $-z^2$. So:

Thus the integral is:

3. Integrate term by term. The first part is easy: $\int z^2\,dz = \frac{z^3}{3}$. For the second part, notice that the derivative of $1+z^3$ is $3z^2$, which is almost the numerator $z^2$. So we adjust:

Therefore:

$4 \int \left( z^2 - \frac{z^2}{1+z^3} \right) dz = 4 \left( \frac{z^3}{3} - \frac{1}{3} \log|1+z^3| \right) + C = \frac{4}{3} \left( z^3 - \log|1+z^3| \right) + C.$ …

Method: Clearing fractional powers with $x=z^{k}$

Use this when the integrand has several fractional powers of $x$ (like $x^{1/2}$ and $x^{3/4}$). Substitute $x=z^{k}$ where $k$ is the least common multiple of the denominators, turning all powers into integers.

Steps

Step 1: Find the LCM of the fractional denominators.

For $x^{1/2}$ and $x^{3/4}$, the denominators are $2$ and $4$, so $k=4$; set $x=z^4$.

Step 2: Convert every power and the differential.

Step 3: Rewrite as a rational function in $z$. …

Common Mistakes

Mistake 1: Choosing the wrong $k$.

Why it's wrong: using $x=z^2$ leaves $x^{3/4}=z^{3/2}$, still fractional; only $k=4$ clears both. Correct approach: take the LCM of all denominators ($2,4\Rightarrow4$).

Mistake 2: Forgetting $dx=4z^3\,dz$.

Why it's wrong: $x=z^4$ gives $dx=4z^3\,dz$; dropping it loses a factor and the $\tfrac43$ coefficient. Correct approach: differentiate the substitution fully. …