Q.Find $\int \dfrac{(x^4 - x)^{1/4}}{x^5}\, dx$

Concept: U Substitution — the numerator’s structure suggests setting $u = 1 - \frac{1}{x^3}$ to simplify the radical.

Step 1: Rewrite the integrand.

Factor $x^4$ inside the fourth root:

$(x^4 - x)^{1/4} = \left[x^4\left(1 - \frac{1}{x^3}\right)\right]^{1/4} = x \left(1 - \frac{1}{x^3}\right)^{1/4}$.

Thus the integral becomes

$\int \frac{x \left(1 - \frac{1}{x^3}\right)^{1/4}}{x^5}\, dx = \int x^{-4} \left(1 - \frac{1}{x^3}\right)^{1/4} dx$.

Step 2: Substitute $u = 1 - \frac{1}{x^3}$. Then $du = \frac{3}{x^4}\, dx$, so $x^{-4} dx = \frac{du}{3}$.

The integral becomes $\int u^{1/4} \cdot \frac{du}{3} = \frac{1}{3} \int u^{1/4}\, du$.

Step 3: Integrate:

$\frac{1}{3} \cdot \frac{u^{5/4}}{5/4} = \frac{4}{15} u^{5/4} + C$.

Step 4: Back-substitute $u = 1 - \frac{1}{x^3}$:

$\frac{4}{15} \left(1 - \frac{1}{x^3}\right)^{5/4} + C$.

Why This Approach Works

When you see an expression like $(x^4 - x)^{1/4}$, your first instinct might be to factor something out. Notice that $x^4 - x = x(x^3 - 1)$. The fourth root then becomes $x^{1/4}(x^3 - 1)^{1/4}$. But the denominator is $x^5$, so the overall power of $x$ in the numerator is $1/4$ from the root, and dividing by $x^5$ gives $x^{1/4 - 5} = x^{-19/4}$. That’s messy.

A better insight: factor $x^4$ out of the bracket instead. Write $x^4 - x = x^4\left(1 - \frac{1}{x^3}\right)$. Then the fourth root becomes $x \left(1 - \frac{1}{x^3}\right)^{1/4}$. Now the integrand is:

This is much cleaner. The denominator $x^4$ suggests that a substitution involving $1/x^3$ might work, because its derivative will bring down a factor of $1/x^4$.

Step-by-Step Solution

1. Rewrite the integrand Factor $x^4$ from $(x^4 - x)$:

Then

The integral becomes:

2. Choose a substitution Let $t = 1 - \frac{1}{x^3}$. Then differentiate:

Notice that $\frac{1}{x^4} dx$ appears in our integral. So we can write:

3. Transform the integral Substituting $t$ and $dt$:

4. Integrate with respect to $t$ Using the power rule:

5. Substitute back Recall $t = 1 - \frac{1}{x^3}$. So:

Method: Factor out the dominant power, then substitute

Use this for integrands like $\dfrac{(x^n-x)^{1/k}}{x^m}$ where a root of a polynomial sits over a power of $x$. Pulling the highest power of $x$ out of the root exposes a clean inner function whose derivative already appears.

Steps

Step 1: Factor the largest power of $x$ out from inside the root.

Write $x^4-x=x^4\!\left(1-\tfrac{1}{x^3}\right)$ so that

Choosing the largest power (not $x$ itself) is what leaves a bracket of the form $1-\tfrac{1}{x^3}$, whose derivative is simple.

Step 2: Simplify the whole integrand.

Cancel the freed power of $x$ against the denominator so the integral reduces to

Step 3: Substitute $u$ = the bracket.

Let $u=1-\tfrac{1}{x^3}$. Then $du=\tfrac{3}{x^4}\,dx$, so $\tfrac{1}{x^4}\,dx=\tfrac{du}{3}$ — exactly the leftover factor. The integral becomes $\tfrac13\int u^{1/4}\,du$.

Step 4: Integrate by the power rule and back-substitute.

$\int u^{1/4}\,du=\tfrac45 u^{5/4}$, giving $\tfrac{4}{15}u^{5/4}+C$; replace $u$ by $1-\tfrac{1}{x^3}$.

Common Mistakes

Mistake 1: Factoring out $x$ instead of $x^4$.

Why it's wrong: writing $x^4-x=x(x^3-1)$ gives $(x^4-x)^{1/4}=x^{1/4}(x^3-1)^{1/4}$, and dividing by $x^5$ leaves the ugly power $x^{-19/4}$ with no clean substitution. Correct approach: factor the largest power ($x^4$) so the bracket becomes $1-\tfrac{1}{x^3}$.

Mistake 2: Dropping the $\tfrac13$ from $du$.

Why it's wrong: $u=1-\tfrac{1}{x^3}\Rightarrow du=\tfrac{3}{x^4}\,dx$, so $\tfrac{1}{x^4}\,dx=\tfrac{du}{3}$, not $du$. Missing the $\tfrac13$ triples the coefficient. Correct approach: solve $du$ for the exact factor appearing in the integral.

Mistake 3: Power-rule slip on $u^{1/4}$.

Why it's wrong: $\int u^{1/4}\,du=\dfrac{u^{5/4}}{5/4}=\tfrac45 u^{5/4}$; combined with $\tfrac13$ this is $\tfrac{4}{15}$, a value students frequently miscompute. Correct approach: add $1$ to $\tfrac14$ to get $\tfrac54$ and divide by it.