Q.Evaluate the integral using substitution $\int_{-1}^{1}\frac{dx}{x^2+2x+5}$

Concept: U Substitution — completing the square in the denominator to match the standard arctangent form.

First, complete the square:

So the integral becomes

Let $u = x+1$, so $du = dx$. When $x = -1$, $u = 0$; when $x = 1$, $u = 2$. The integral is

This is a standard form: $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$. Here $a = 2$, so …

We evaluate $\displaystyle\int_{-1}^{1}\frac{dx}{x^2+2x+5}$.

1. Complete the square.

2. Substitute $u=x+1$, so $du=dx$. The limits change: $x=-1\Rightarrow u=0$ and $x=1\Rightarrow u=2$.

$\int_{0}^{2}\frac{du}{u^2+4}$ …

Method: Complete the square to reach the arctangent standard form

An irreducible quadratic denominator (negative discriminant) becomes $u^2+a^2$ after completing the square, giving an inverse-tangent integral.

Steps

Step 1: Complete the square.

Write $x^2+bx+c=(x+\tfrac{b}{2})^2 + \big(c-\tfrac{b^2}{4}\big)$; when $c-\tfrac{b^2}{4}=a^2>0$ the denominator is $u^2+a^2$. …

Common Mistakes

Mistake 1: Trying partial fractions on an irreducible quadratic.

Why it's wrong: $x^2+2x+5$ has negative discriminant, so it has no real linear factors — partial fractions fail. Correct approach: complete the square to $(x+1)^2+4$ and use the arctan form.

Mistake 2: Not shifting the limits after $u=x+1$. …