Q.If $\int \dfrac{dx}{(x+2)(x^2+1)} = a\log|1+x^2| + b\tan^{-1}x + \dfrac{1}{5}\log|x+2| + C$, then
(A) $a=-\dfrac{1}{10},\ b=-\dfrac{2}{5}$
(B) $a=\dfrac{1}{10},\ b=-\dfrac{2}{5}$
(C) $a=-\dfrac{1}{10},\ b=\dfrac{2}{5}$
(D) $a=\dfrac{1}{10},\ b=\dfrac{2}{5}$

Concept: Partial Fraction Decomposition

We decompose

Multiplying through:

Step 1 – Find $A$

Put $x = -2$:

Step 2 – Compare coefficients

Expand:

Equating:

• $x^2$: $A+B = 0 \Rightarrow B = -\frac{1}{5}$
• $x$: $2B+C = 0 \Rightarrow C = \frac{2}{5}$
• Constant: $A+2C = 1$ checks out.

Step 3 – Integrate

The second integral splits: …

The problem gives us the answer structure before we start — that’s a huge clue. The integral of a rational function like $\frac{1}{(x+2)(x^2+1)}$ is almost always found by partial fraction decomposition. The denominator is already factored: one linear factor $(x+2)$ and one irreducible quadratic $(x^2+1)$. The form on the right tells us the decomposition will produce three pieces: a term giving $\log|x+2|$, a term giving $\log|1+x^2|$, and a term giving $\tan^{-1}x$. Our job is to find the constants $a$ and $b$ that make the equality hold.

Let’s work through it.

1. Set up the partial fractions. Since the denominator has a linear factor and an irreducible quadratic, we write:

The numerator over $x^2+1$ is linear ($Bx+C$) because the quadratic doesn’t factor further over the reals. This is the standard form.

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

Expand carefully:

Group like powers of $x$:

3. Equate coefficients. The left side is $1$, which we can think of as $0x^2 + 0x + 1$. So:

From the first equation, $B = -A$. Substitute into the second: $2(-A) + C = 0 \Rightarrow C = 2A$.

Now put $C = 2A$ into the third: $A + 2(2A) = 1 \Rightarrow A + 4A = 1 \Rightarrow 5A = 1 \Rightarrow A = \frac{1}{5}$.

Then $B = -\frac{1}{5}$ and $C = \frac{2}{5}$.

So the decomposition is:

4. Integrate term by term.

The first integral is straightforward:

For the second, split the numerator:

The first of these is a simple substitution: let $u = x^2+1$, so $du = 2x\,dx$, and $-\frac{1}{5}x\,dx = -\frac{1}{10}\,du$. Hence: …

Method: Linear factor times an irreducible quadratic

Use when the denominator is (linear)$\times$(irreducible quadratic), e.g. $(x+p)(x^2+q)$; the quadratic contributes a linear numerator.

Steps

Step 1: Set up the mixed template.

Step 2: Solve for $A$, $B$, $C$.

Get $A$ by cover-up (put $x=-p$); get $B,C$ by comparing the coefficients of $x^2$, $x$, and the constant. …

Common Mistakes

Mistake 1: Forgetting the $\frac12$ when integrating $\frac{x}{x^2+1}$.

Why it's wrong: with $u=x^2+1$, $x\,dx=\frac12\,du$, so $-\frac15\int\frac{x}{x^2+1}\,dx=-\frac{1}{10}\log|x^2+1|$. Correct approach: this gives $a=-\frac{1}{10}$, not $-\frac15$.

Mistake 2: Using a constant numerator over $x^2+1$. …