Q.Integrate the following function: $\int \frac{dx}{x^2 + 2x + 2}$ equals (A) $x \tan^{-1} (x + 1) + C$ (B) $\tan^{-1} (x + 1) + C$ (C) $(x + 1) \tan^{-1} x + C$ (D) $\tan^{-1} x + C$

Concept: Integration by completing the square.

Step 1: Complete the square in the denominator:

Step 2: The integral becomes

$\int \frac{dx}{(x+1)^2 + 1}.$ …

When you see a quadratic denominator that doesn't factor nicely into linear terms, the standard play is to complete the square. Why? Because the integral $\int \frac{dx}{x^2 + a^2}$ is a known form — it gives an inverse tangent. The trick is to force the denominator into that shape: a perfect square plus a constant.

Here, $x^2 + 2x + 2$ is almost $(x+1)^2$, but not quite. Let's see why that matters.

1. Complete the square Take $x^2 + 2x$. Half of 2 is 1, so $(x+1)^2 = x^2 + 2x + 1$. Our denominator has $+2$ instead of $+1$, so:

The integral becomes:

2. Substitute to match the standard form Let $u = x+1$, so $du = dx$. Then:

This is exactly $\int \frac{du}{u^2 + a^2}$ with $a = 1$.

3. Apply the formula With $a = 1$, we get:

4. Back-substitute Replace $u$ with $x+1$: $\tan^{-1}(x+1) + C.$ …

Method: Integral of $\dfrac{1}{\text{quadratic}}$ with no real roots (arctan form)

Use this for $\displaystyle\int\frac{dx}{ax^2+bx+c}$ when the quadratic does not factor over the reals (discriminant $b^2-4ac<0$). Completing the square turns it into the standard arctangent integral.

Steps

Step 1: Complete the square in the denominator.

Write $ax^2+bx+c = a\left[(x+h)^2 + k^2\right]$ where $h=\dfrac{b}{2a}$ and $k^2>0$ because the discriminant is negative.

Step 2: Substitute $u=x+h$ (so $du=dx$) to reach the standard shape $\displaystyle\int\frac{du}{u^2+k^2}$.

Step 3: Apply the standard formula. …

Common Mistakes

Mistake 1: Ignoring the shift and answering $\tan^{-1}x$ (distractor D).

Why it's wrong: the denominator is $(x+1)^2+1$, centred at $x=-1$, not at $0$; the antiderivative is $\tan^{-1}(x+1)$. Correct approach: complete the square, substitute $u=x+1$, and keep the shift in the final answer.

Mistake 2: Introducing a stray factor of $x$ (distractor A) or of $(x+1)$ (distractor C).

Why it's wrong: $\int\frac{du}{u^2+1}=\tan^{-1}u$ has no polynomial multiplier out front; options like $x\tan^{-1}(x+1)$ or $(x+1)\tan^{-1}x$ come from mis-remembering the formula. Correct approach: use $\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\frac ua$ exactly, with $a=1$ here. …