Key Concepts and Formulas

• Completing the Square: Transforming a quadratic expression $ax^2+bx+c$ into the form $a(x-h)^2+k$. For $x^2+bx$, it becomes $(x + b/2)^2 - (b/2)^2$.
• Trigonometric Substitution: For integrals involving $\sqrt{a^2+x^2}$, $\sqrt{a^2-x^2}$, or $\sqrt{x^2-a^2}$, specific substitutions are used:
  • $x = a\tan\theta$ for $\sqrt{a^2+x^2}$
  • $x = a\sin\theta$ for $\sqrt{a^2-x^2}$
  • $x = a\sec\theta$ for $\sqrt{x^2-a^2}$
• Definite Integral Substitution Rule: When changing the variable in a definite integral, the limits of integration must also be changed to correspond to the new variable.
• Trigonometric Identity: $\tan^2\theta + 1 = \sec^2\theta$.

Step-by-Step Solution

Step 1: Simplify the Denominator by Completing the Square

The given integral is $I = \int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}}$. The expression in the denominator, $x^2 - 2x + 4$, can be simplified by completing the square. We focus on the $x^2 - 2x$ part. Half of the coefficient of $x$ is $-1$, and squaring it gives $1$. $x^2 - 2x + 4 = (x^2 - 2x + 1) - 1 + 4 = (x - 1)^2 + 3$ This transforms the integral into: $I = \int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}}$ This form, $(u^2 + a^2)^{3/2}$ with $u = x-1$ and $a = \sqrt{3}$, suggests a trigonometric substitution.

Step 2: Apply Trigonometric Substitution

For an expression of the form $(u^2 + a^2)^{3/2}$, we use the substitution $u = a\tan\theta$. Here, $u = x-1$ and $a = \sqrt{3}$. Let $x - 1 = \sqrt{3} \tan\theta$. Differentiating both sides with respect to $\theta$ to find $dx$: $dx = \sqrt{3} \sec^2\theta \, d\theta$

Step 3: Change the Limits of Integration

Since this is a definite integral, we must change the limits from $x$ values to $\theta$ values.

• When $x = 1$: $1 - 1 = \sqrt{3} \tan\theta \implies 0 = \sqrt{3} \tan\theta \implies \tan\theta = 0$. Thus, $\theta = 0$.
• When $x = 2$: $2 - 1 = \sqrt{3} \tan\theta \implies 1 = \sqrt{3} \tan\theta \implies \tan\theta = \frac{1}{\sqrt{3}}$. Thus, $\theta = \frac{\pi}{6}$. The new limits of integration are from $0$ to $\frac{\pi}{6}$.

Step 4: Substitute and Simplify the Integral

Substitute $x-1 = \sqrt{3}\tan\theta$, $dx = \sqrt{3}\sec^2\theta\,d\theta$, and the new limits into the integral: $I = \int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left[ {{{\left( {\sqrt 3 \tan\theta} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}}$ Simplify the term inside the bracket: $(\sqrt{3}\tan\theta)^2 + 3 = 3\tan^2\theta + 3 = 3(\tan^2\theta + 1)$ Using the identity $\tan^2\theta + 1 = \sec^2\theta$: $3(\tan^2\theta + 1) = 3\sec^2\theta$ Now, substitute this back into the denominator's bracket: ${\left[ {3{{\sec }^2}\theta } \right]^{{3 \over 2}}} = {3^{3/2}} (\sec^2\theta)^{3/2} = 3\sqrt{3} \sec^3\theta$ The integral becomes: $I = \int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}}$ Cancel common terms: $I = \int\limits_0^{{\pi \over 6}} {{1 \,d\theta } \over {3\sec \theta }} = {1 \over 3} \int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta }$

Step 5: Evaluate the Definite Integral

Integrate $\cos\theta$ with respect to $\theta$: $I = {1 \over 3} \left[ {\sin \theta } \right]_0^{{\pi \over 6}}$ Apply the limits: $I = {1 \over 3} \left( {\sin \left( {{\pi \over 6}} \right) - \sin \left( 0 \right)} \right)$ $I = {1 \over 3} \left( {1 \over 2} - 0 \right) = {1 \over 3} \times {1 \over 2} = {1 \over 6}$

Step 6: Solve for k

We are given that the integral equals $\frac{k}{k+5}$. $\frac{k}{k+5} = \frac{1}{6}$ Cross-multiply: $6k = 1(k+5)$ $6k = k+5$ $5k = 5$ $k = 1$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when completing the square and simplifying algebraic expressions involving powers.
• Limit Conversion: Always convert the limits of integration when performing a substitution in a definite integral.
• Trigonometric Simplification: Ensure correct application of trigonometric identities and accurate simplification of powers of trigonometric functions (e.g., $(\sec^2\theta)^{3/2} = \sec^3\theta$).

Summary

The problem requires evaluating a definite integral containing a quadratic expression raised to a fractional power. The standard approach involves completing the square to identify the form of the integrand, followed by a trigonometric substitution. The substitution $x-1 = \sqrt{3}\tan\theta$ was used, and the limits of integration were adjusted accordingly. After simplifying the integral using trigonometric identities, it was evaluated to $\frac{1}{6}$. Equating this result to the given expression $\frac{k}{k+5}$ allowed us to solve for $k$.

The final answer is $\boxed{1}$.