Q.Integrate the following function: $(x^3-1)^{1/3} x^5$

Concept: U Substitution — we set $u = x^3 - 1$ to simplify the power and then rewrite $x^5$ in terms of $u$.

Let $u = x^3 - 1$. Then $du = 3x^2\,dx$, so $x^2\,dx = \frac{du}{3}$.

Also $x^3 = u + 1$, hence $x^5 = x^3 \cdot x^2 = (u+1)x^2$.

Substitute:

Integrate: …

1. Substitute. Let $u = x^3 - 1$, so $du = 3x^2\,dx$, i.e. $x^2\,dx = \tfrac13\,du$, and $x^3 = u+1$.

2. Rewrite $x^5\,dx$.

3. Transform the integral.

4. Integrate by the power rule.

5. Factor and back-substitute. …

Method: Substitute, then reuse the substitution to convert the extra powers

Use this when the power of $x$ outside is higher than the derivative supplies — e.g. $(x^3-1)^{1/3}x^5$: substitute $u=x^3-1$, split $x^5=x^3\cdot x^2$, and re-express the leftover $x^3$ as $u+1$.

Steps

Step 1: Substitute and peel off $x^2\,dx$ for $du$.

Let $u=x^3-1$, so $du=3x^2\,dx$. Write $x^5\,dx=x^3\cdot x^2\,dx=x^3\cdot\dfrac{du}{3}$.

Step 2: Replace the remaining $x^3$ using the substitution.

From $u=x^3-1$ we get $x^3=u+1$, so the integral becomes …

Common Mistakes

Mistake 1: Not splitting $x^5$ into $x^3\cdot x^2$.

Why it's wrong: only $x^2\,dx$ matches $du=3x^2\,dx$; the remaining $x^3$ must be handled separately. Correct approach: write $x^5\,dx=x^3\cdot x^2\,dx$ and peel off $x^2\,dx$ for $du$.

Mistake 2: Forgetting to re-express the leftover $x^3$ as $u+1$.

Why it's wrong: with $u=x^3-1$, $x^3=u+1$; leaving $x^3$ un-substituted leaves the integral in two variables. Correct approach: replace $x^3$ by $u+1$ to get $\tfrac13\int u^{1/3}(u+1)\,du$. …