Q.Integrate the function $\frac{1}{x-x^3}$

Concept: Partial Fraction Decomposition — splitting a rational function into simpler fractions that integrate to logarithms.

First, factor the denominator:

Now decompose into partial fractions:

Solving gives $A = 1$, $B = \frac{1}{2}$, $C = -\frac{1}{2}$.

Integrate term by term:

Combine the logarithms:

The integral $\int \frac{1}{x-x^3}\,dx$ looks simple, but the denominator is a cubic — and that’s the clue. Whenever you see a polynomial in the denominator that factors nicely, partial fractions are your best friend. The idea is to break a complicated fraction into a sum of simpler ones, each of which integrates to a logarithm (or a simple rational function).

Here, $x - x^3 = x(1 - x^2) = x(1-x)(1+x)$. So we have three distinct linear factors. That means we can write:

for some constants $A, B, C$. Once we find them, integration becomes straightforward.

Let’s work through it step by step.

1. Set up the decomposition. Multiply both sides by the denominator $x(1-x)(1+x)$ to clear fractions:

Notice that $(1-x)(1+x) = 1 - x^2$, so the first term is $A(1-x^2)$. The other two expand as $Bx + Bx^2$ and $Cx - Cx^2$.

2. Expand and collect like terms.

Group powers of $x$:

• Constant term: $A$
• $x$ term: $(B + C)x$
• $x^2$ term: $(-A + B - C)x^2$

So we have:

3. Equate coefficients. The left side is $1 + 0\cdot x + 0\cdot x^2$. Therefore:

From $A=1$, the third equation becomes $-1 + B - C = 0$, i.e. $B - C = 1$.

Together with $B + C = 0$, we solve:

• Adding: $2B = 1 \Rightarrow B = \frac{1}{2}$
• Then $C = -\frac{1}{2}$

So $A = 1$, $B = \frac{1}{2}$, $C = -\frac{1}{2}$.

4. Rewrite the integral.

5. Integrate term by term.

   • $\int \frac{1}{x}\,dx = \log|x| + C_1$
   • $\int \frac{1/2}{1-x}\,dx = \frac{1}{2} \int \frac{1}{1-x}\,dx = -\frac{1}{2} \log|1-x| + C_2$ (because the derivative of $1-x$ is $-1$)
   • $\int -\frac{1/2}{1+x}\,dx = -\frac{1}{2} \log|1+x| + C_3$

   Combine constants into a single $C$:

6. Simplify using logarithm properties. Factor the $-\frac{1}{2}$:

The sum of logs is the log of the product:

And $(1-x)(1+x) = 1 - x^2$, so:

Combine into a single logarithm:

Method: Partial fractions over distinct linear factors

Use this for $\displaystyle\int\frac{P(x)}{Q(x)}\,dx$ where $Q$ factors completely into different linear pieces and $\deg P<\deg Q$; each piece integrates to a logarithm.

Steps

Step 1: Factor the denominator completely.

Pull out every linear factor. A denominator that looks cubic often hides a common factor — always check for one before decomposing.

Step 2: Write one term per factor with unknown constants.

Step 3: Solve for the constants (cover-up shortcut).

To get the constant over $(x-r_i)$, delete that factor and evaluate the rest at $x=r_i$. This is faster and less error-prone than expanding and matching coefficients.

Step 4: Integrate term by term.

Each $\int\frac{k}{x-r}\,dx=k\log|x-r|$. Keep the absolute value, and watch the chain-rule sign when the factor is $(1-x)$ rather than $(x-1)$: $\int\frac{dx}{1-x}=-\log|1-x|$.

Common Mistakes

Mistake 1: Not factoring the denominator fully.

Why it's wrong: $x-x^3=x(1-x)(1+x)$ has three linear factors; stopping at $x(1-x^2)$ (or missing the $x$) blocks the decomposition. Correct approach: factor completely into $x(1-x)(1+x)$ before setting up partial fractions.

Mistake 2: Sign error integrating $\dfrac{1}{1-x}$.

Why it's wrong: because $\tfrac{d}{dx}(1-x)=-1$, $\int\dfrac{dx}{1-x}=-\log|1-x|$, not $+\log|1-x|$. Correct approach: apply the chain-rule sign for factors of the form $(1-x)$.

Mistake 3: Dropping the absolute values.

Why it's wrong: the integrand is undefined at $x=0,\pm1$, so the antiderivative must use $\log|\cdot|$. Correct approach: keep $\log|\cdot|$ throughout.