Question

Evaluate the definite integral:
\[ \int_{0}^{1} \frac{1}{x^2 + 2x + 3} \, dx \]

Hint:

When completing the square for $x^2 + bx + c$, always rewrite it as $\left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right)$. Keeping this mental shortcut handy makes working through algebraic steps much faster during exams.

Answer 1

Concept: This problem requires integrating a rational function with a quadratic denominator. Since the denominator cannot be factored easily into linear components over the real numbers, we use the method of completing the square to transform it into the standard form: \[ \int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C \]

Step 1: Completing the square for the denominator expression
Consider the quadratic function in the denominator: $x^2 + 2x + 3$. To rewrite this as a perfect square plus a constant, take half of the coefficient of $x$ (which is $\frac{2}{2} = 1$), square it ($1^2 = 1$), and then add and subtract it: \[ x^2 + 2x + 3 = (x^2 + 2x + 1) - 1 + 3 \] \[ x^2 + 2x + 3 = (x + 1)^2 + 2 \] We can express the constant 2 as a square: $2 = (\sqrt{2})^2$. Thus, the quadratic terms become: \[ x^2 + 2x + 3 = (x + 1)^2 + (\sqrt{2})^2 \]

Step 2: Rewriting the definite integral with new limits
Let us substitute our rewritten expression into the integral $I$: \[ I = \int_{0}^{1} \frac{1}{(x + 1)^2 + (\sqrt{2})^2} \, dx \] Let us apply a variable substitution to simplify the expression: \[ \text{Let } u = x + 1 \implies du = dx \] Now, we must update the limits of integration according to this substitution:
• Lower limit: When $x = 0 \implies u = 0 + 1 = 1$
• Upper limit: When $x = 1 \implies u = 1 + 1 = 2$ Substituting these values gives: \[ I = \int_{1}^{2} \frac{1}{u^2 + (\sqrt{2})^2} \, du \]

Step 3: Evaluating the antiderivative
Using the standard integration formula with $a = \sqrt{2}$: \[ I = \left[ \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) \right]_{1}^{2} \] We can factor out the constant scalar value $\frac{1}{\sqrt{2}}$: \[ I = \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) \right]_{1}^{2} \]

Step 4: Substituting the limits of integration
Applying the Fundamental Theorem of Calculus by evaluating at the upper limit (2) and subtracting the value at the lower limit (1): \[ I = \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left(\frac{2}{\sqrt{2}}\right) - \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \right] \] Simplifying the fraction inside the first inverse tangent term ($\frac{2}{\sqrt{2}} = \sqrt{2}$), we get our final exact value: \[ I = \frac{1}{\sqrt{2}} \left[ \tan^{-1}(\sqrt{2}) - \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \right] \]