Partial fractions — decompose and integrate (2x+1)/((x+1)(x²+1))

Question

Decompose $\frac{2x+1}{(x+1)(x^2+1)}$ into partial fractions and hence evaluate:

\int \frac{2x+1}{(x+1)(x^2+1)} \, dx

(CBSE 2024)

Solution — Step by Step

Since the denominator has a linear factor $(x+1)$ and an irreducible quadratic $(x^2+1)$ :

\frac{2x+1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+1}

Multiply both sides by $(x+1)(x^2+1)$ :

2x + 1 = A(x^2 + 1) + (Bx + C)(x + 1)

Put $x = -1$ : $2(-1) + 1 = A(1 + 1) + 0$

$-1 = 2A \implies A = -\frac{1}{2}$

Compare $x^2$ coefficients: $0 = A + B \implies B = -A = \frac{1}{2}$

Compare constant terms: $1 = A + C \implies C = 1 - (-\frac{1}{2}) = \frac{3}{2}$

So: $\frac{2x+1}{(x+1)(x^2+1)} = \frac{-1/2}{x+1} + \frac{x/2 + 3/2}{x^2+1}$

\int \frac{2x+1}{(x+1)(x^2+1)} \, dx = -\frac{1}{2}\int\frac{dx}{x+1} + \frac{1}{2}\int\frac{x}{x^2+1} \, dx + \frac{3}{2}\int\frac{dx}{x^2+1}

First integral: $-\frac{1}{2}\ln|x+1|$

Second integral: $\frac{1}{2} \cdot \frac{1}{2}\ln(x^2+1) = \frac{1}{4}\ln(x^2+1)$

(using substitution $u = x^2 + 1$ )

Third integral: $\frac{3}{2}\tan^{-1}(x)$

\boxed{-\frac{1}{2}\ln|x+1| + \frac{1}{4}\ln(x^2+1) + \frac{3}{2}\tan^{-1}(x) + C}

Why This Works

Partial fractions break a complex fraction into simpler pieces that we know how to integrate. The key insight is that any rational function (polynomial over polynomial) can be decomposed into:

$\frac{A}{ax + b}$ for linear factors → integrates to $\ln$
$\frac{Bx + C}{x^2 + px + q}$ for irreducible quadratic factors → integrates to $\ln$ and $\tan^{-1}$

The irreducible quadratic $x^2 + 1$ needs a linear numerator $Bx + C$ (not just a constant $B$ ), because a quadratic denominator can absorb a first-degree numerator.

Alternative Method — Cover-up for A, then compare

Use the cover-up rule for $A$ : cover $(x+1)$ in the original fraction and substitute $x = -1$ :

A = \frac{2(-1) + 1}{(-1)^2 + 1} = \frac{-1}{2} = -\frac{1}{2}

Then for $B$ and $C$ , substitute any two convenient values (say $x = 0$ and $x = 1$ ) into the identity and solve.

When splitting $\frac{Bx + C}{x^2 + 1}$ for integration, always separate it as $\frac{Bx}{x^2+1} + \frac{C}{x^2+1}$ . The first part uses substitution ( $u = x^2 + 1$ ), the second is a standard $\tan^{-1}$ form. This split is the key step that makes the integration straightforward.

Common Mistake

Students write the partial fraction for the irreducible quadratic as $\frac{B}{x^2+1}$ (constant numerator) instead of $\frac{Bx + C}{x^2+1}$ (linear numerator). With just a constant, you won’t have enough unknowns to match all coefficients, and the decomposition will fail. Always use a linear numerator $Bx + C$ over an irreducible quadratic.

Question

Solution — Step by Step

Why This Works

Alternative Method — Cover-up for A, then compare

Common Mistake

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