Q.Prove that $\int_{1}^{3}\frac{dx}{x^2(x+1)}=\frac{2}{3}+\log\frac{2}{3}$

Concept: Partial Fraction Decomposition — splitting the rational function into simpler fractions that integrate directly.

First, decompose the integrand:

Multiplying through by $x^2(x+1)$ gives:

Comparing coefficients (or substituting convenient $x$ values):

• For $x=0$: $1 = B \implies B = 1$.
• For $x=-1$: $1 = C \implies C = 1$.
• For $x=1$: $1 = A(2) + 1(2) + 1(1) \implies 2A + 3 = 1 \implies A = -1$.

Thus:

Now integrate term by term from $1$ to $3$:

$\int_{1}^{3} \left( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \right) dx = \left[ -\log|x| - \frac{1}{x} + \log|x+1| \right]_{1}^{3}.$ …

We need to evaluate

The integrand is a rational function where the denominator is already factored. The key idea is Partial Fraction Decomposition — we break the complicated fraction into a sum of simpler fractions that we can integrate directly using basic formulas (like $\int x^n dx$ and $\int \frac{dx}{x+a}$).

Why does this work? Because the denominator has a repeated linear factor $x^2$ and a distinct linear factor $(x+1)$. Each factor contributes its own term in the decomposition, with the repeated factor requiring terms for each power.

Let’s go step by step.

1. Set up the partial fractions Since $x^2$ is a repeated factor (power 2), we write:

Here $A$, $B$, $C$ are constants to be found.

2. Clear the denominator Multiply both sides by $x^2(x+1)$:

Expand:

Group like powers of $x$:

3. Equate coefficients

   The left side is $0x^2 + 0x + 1$. So:

   • Coefficient of $x^2$: $A + C = 0$
   • Coefficient of $x$: $A + B = 0$
   • Constant term: $B = 1$

   From $B = 1$, then $A + 1 = 0 \Rightarrow A = -1$.

   Then $A + C = 0 \Rightarrow -1 + C = 0 \Rightarrow C = 1$.

   So:

4. Integrate term by term Now:

Integrate each:

• $\int -\frac{1}{x} dx = -\log|x|$
• $\int \frac{1}{x^2} dx = \int x^{-2} dx = -x^{-1} = -\frac{1}{x}$
• $\int \frac{1}{x+1} dx = \log|x+1|$

So an antiderivative is:

Combine the logs: …

Method: Partial fractions with a repeated factor $x^{2}$

Use this for $\dfrac{1}{x^{2}(x+c)}$: the squared factor $x^2$ requires both a $\dfrac{A}{x}$ and a $\dfrac{B}{x^2}$ term.

Steps

Step 1: Set up the decomposition with all powers.

Step 2: Clear denominators and solve.

Match coefficients (or substitute $x=0$ for $B$, $x=-1$ for $C$) to find $A,B,C$. …

Common Mistakes

Mistake 1: Omitting the $\dfrac{B}{x^2}$ term.

Why it's wrong: the repeated factor $x^2$ needs both $\dfrac{A}{x}$ and $\dfrac{B}{x^2}$; leaving one out makes the decomposition fail. Correct approach: include a term for each power.

Mistake 2: Integrating $\dfrac{1}{x^2}$ as a logarithm.

Why it's wrong: $\int x^{-2}\,dx=-\dfrac1x$, not $\log|x|$. Correct approach: use the power rule for the $x^{-2}$ term. …