Q.Integrate the function $\frac{x^2}{1-x^6}$

Concept: Partial Fraction Decomposition — we factor the denominator and split the rational function into simpler fractions that integrate to inverse trigonometric functions.

Step 1: Factor the denominator.

$1 - x^6 = (1 - x^3)(1 + x^3) = (1 - x)(1 + x + x^2)(1 + x)(1 - x + x^2)$.

Step 2: Use a substitution to simplify.

Let $u = x^3$, so $du = 3x^2 dx$. Then

Step 3: Integrate. …

Note $1-x^6=1-(x^3)^2$ and the numerator $x^2$ is a constant multiple of the derivative of $x^3$, so substitute $u=x^3$.

Substitution: $u=x^3\Rightarrow du=3x^2\,dx\Rightarrow x^2\,dx=\tfrac13\,du$:

Standard integral:

so …

Method: Substitution to a $\dfrac{1}{1-u^2}$ Standard Form

Use this when the numerator is a constant multiple of the derivative of an inner expression and the denominator becomes $1-u^2$ (or $1+u^2$) after substitution.

Steps

Step 1: Spot the inner function and its derivative.

Note $1-x^6=1-(x^3)^2$ and the numerator $x^2$ is $\tfrac13$ of $\dfrac{d}{dx}x^3$. So set $u=x^3$, $du=3x^2\,dx$, i.e. $x^2\,dx=\tfrac13\,du$.

Step 2: Reduce to the standard integral. …

Common Mistakes

Mistake 1: Confusing $\dfrac{1}{1-u^2}$ with $\dfrac{1}{1+u^2}$.

Why it's wrong: $\frac{1}{1-u^2}$ gives a logarithm, while $\frac{1}{1+u^2}$ gives $\tan^{-1}u$. Correct approach: check the sign of the $u^2$ term — a minus means the log formula.

Mistake 2: Dropping the $\tfrac13$ from $x^2\,dx=\tfrac13\,du$.

Why it's wrong: $du=3x^2\,dx$, so the numerator is only one-third of $du$. Correct approach: solve $du$ for $x^2\,dx$ before substituting. …