Michael Bowen's VC Course Pages

Answers

Math V21B Section 7.4

Introduction

Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient.

Problem 37

The original problem is $$I = \int {\frac{{{x^2} - 3x + 7}}{{{{({x^2} - 4x + 6)}^2}}}\;dx} $$

Since neither of the quadratic expressions is factorable, and since the degree of the denominator exceeds that of the numerator, it seems that the partial-fractions approach might work. We may write the integrand in two ways (left and right sides of the equation): $$\frac{{{x^2} - 3x + 7}}{{{{\left( {{x^2} - 4x + 6} \right)}^2}}} = \frac{{Ax + B}}{{\left( {{x^2} - 4x + 6} \right)}} + \frac{{Cx + D}}{{{{\left( {{x^2} - 4x + 6} \right)}^2}}}$$ where $A$, $B$, $C$, and $D$ are undetermined coefficients. The LCD is ${\left( {{x^2} - 4x + 6} \right)^2}$, and we clear fractions by multiplying both sides of the equation: $${\left( {{x^2} - 4x + 6} \right)^2}\frac{{{x^2} - 3x + 7}} {{{{\left( {{x^2} - 4x + 6} \right)}^2}}} = {\left( {{x^2} - 4x + 6} \right)^2}\frac{{Ax + B}} {{\left( {{x^2} - 4x + 6} \right)}} + {\left( {{x^2} - 4x + 6} \right)^2}\frac{{Cx + D}} {{{{\left( {{x^2} - 4x + 6} \right)}^2}}}$$ which simplifies to $${x^2} - 3x + 7 = \left( {{x^2} - 4x + 6} \right)\left( {Ax + B} \right) + \left( {Cx + D} \right)$$ or, after distributing the right side and inserting a null $x^3$ term on the left side, $$0{x^3} + 1{x^2} - 3x + 7 = A{x^3} + B{x^2} - 4A{x^2} - 4Bx + 6Ax + 6B + Cx + D$$

Collecting terms on the right side of the equation gives $${x^3}(0) + {x^2}(1) + x( - 3) + (7) = {x^3}(A) + {x^2}(B - 4A) + x( - 4B + 6A + C) + (6B + D)$$ Given that this equality must hold for any arbitrary value of $x$, it necessarily follows that the coefficients of corresponding powers of $x$ must be equal. This gives rise to the $4 \times 4$ linear system $$\begin{array}{*{20}{r}} A&{}&{}&{}&{}&{}&{}& = &0 \\ { - 4A}& + &B&{}&{}&{}&{}& = &1 \\ {6A}& - &{4B}& + &C&{}&{}& = &{ - 3} \\ {}&{}&{6B}&{}&{}& + &D& = &7 \end{array}$$ whose solution $(A,B,C,D) = (0,1,1,1)$ is readily obtained by repeated substitutions.

Replacing $A$, $B$, $C$, and $D$ in the original partial-fraction expansion of the integrand yields $$I = \int {\left[ {\frac{1}{{\left( {{x^2} - 4x + 6} \right)}} + \frac{{x + 1}}{{{{\left( {{x^2} - 4x + 6} \right)}^2}}}} \right]\;dx} = \int {\frac{1}{{\left( {{x^2} - 4x + 6} \right)}}\;dx} + \int {\frac{{x + 1}}{{{{\left( {{x^2} - 4x + 6} \right)}^2}}}\;dx} $$

We first attack the second integral by considering the $u$-substitution $$u = {x^2} - 4x + 6;\;\;\;du = (2x - 4)dx;\;\;\;\frac{{du}}{2} = (x - 2)dx$$ which causes us to contemplate re-writing the numerator $x + 1$ of the second integrand as $(x - 2) + 3$, leading to a further separation of the terms into three integrals: $$I = \int {\frac{1}{{\left( {{x^2} - 4x + 6} \right)}}\;dx} + \int {\frac{{x - 2}}{{{{\left( {{x^2} - 4x + 6} \right)}^2}}}\;dx} + \int {\frac{3}{{{{\left( {{x^2} - 4x + 6} \right)}^2}}}\;dx} $$ This causes the numerator of the second integral to closely resemble the differential $du$, but at the cost of creating yet another integral to solve.

Completing the square in the denominators of the first and third integrands, and implementing the $u$-substitution in the second integral, allows us to rewrite them as $$I = \int {\frac{1}{{{{\left( {x - 2} \right)}^2} + 2}}\;dx} \; + \int {\frac{ \left({\tfrac{{du}}{2}}\right) } {{{u^2}}}} \;\; + \; 3\int {\frac{1}{{{{\left[ {{{\left( {x - 2} \right)}^2} + 2} \right]}^2}}}\;dx} $$

Another $u$-substitution is now indicated for the first and third integrals, in order to re-write the denominators in a form that is compatible with a future trigonometric substitution. However, we have already used the variable $u$ in the second integral, so this time let's use $v$. The substitution is $$v = x - 2;\;\;\;dv = dx;\;\;\;{a^2} = 2;\;\;\;a = \sqrt 2 $$ which gives us a new set of integrals: $$I \; = \; \int {\frac{1}{{{v^2} + {a^2}}}\;dv} \;\; + \;\; \frac{1}{2}\int {\frac{{du}}{{{u^2}}}} \;\; + \;\; 3\int {\frac{1}{{{{\left( {{v^2} + {a^2}} \right)}^2}}}\;dv} $$

The denominators of the first and third integrands now suggest a trigonometric substitution of the form $$v = a\;\tan \;\theta ;\;\;\;dv = a\;{\sec ^2}\theta \;d\theta ;\;\;\;\frac{v}{a} = \tan \;\theta $$ These substitutions, coupled with rewriting the second integrand in the form of a power, lead to $$I \; = \; \int {\frac{1}{{{a^2}\,{{\tan }^2}\theta + {a^2}}}\left( {a\,{{\sec }^2}\theta \;d\theta } \right)} \;\; + \;\; \frac{1}{2}\int {{u^{ - 2}}\;du} \;\; + \;\; 3\int {\frac{1}{{{{\left( {{a^2}\,{{\tan }^2}\theta + {a^2}} \right)}^2}}}\left( {a\,{{\sec }^2}\theta \;d\theta } \right)} $$ which we can simplify to $$I \; = \; \int {\frac{{a\,{{\sec }^2}\theta \;d\theta }}{{{a^2}\left( {{{\tan }^2}\theta + 1} \right)}}} \;\; + \;\; \frac{1}{2}\int {{u^{ - 2}}\;du} \;\; + \;\; 3\int {\frac{{a\,{{\sec }^2}\theta \;d\theta }}{{{{\left[ {{a^2}\left( {{{\tan }^2}\theta + 1} \right)} \right]}^2}}}} $$ $$I \; = \; \frac{1}{a}\int {\frac{{{{\sec }^2}\theta \;d\theta }}{{{{\sec }^2}\theta }}} \;\; + \;\; \frac{1}{2}\int {{u^{ - 2}}\;du} \;\; + \;\; \frac{3}{{{a^3}}}\int {\frac{{{{\sec }^2}\theta \;d\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}} $$ $$I \; = \; \frac{1}{a}\int {d\theta } \;\; + \;\; \frac{1}{2}\int {{u^{ - 2}}\;du} \;\; + \;\; \frac{3}{{{a^3}}}\int {\frac{{d\theta }}{{{{\sec }^2}\theta }}} $$ $$I \; = \; \frac{1}{a}\int {d\theta } \;\; + \;\; \frac{1}{2}\int {{u^{ - 2}}\;du} \;\; + \;\; \frac{3}{{{a^3}}}\int {{{\cos }^2}\theta \;d\theta } $$

We may further simplify the last integrand using the identity ${\cos ^2}\theta = \tfrac{1}{2} + \tfrac{1}{2}\;\cos \,2\theta $: $$I \; = \; \frac{1}{a}\int {d\theta } \;\; + \;\; \frac{1}{2}\int {{u^{ - 2}}\;du} \;\; + \;\; \frac{3}{{{a^3}}}\int {\left( {\tfrac{1}{2} + \tfrac{1}{2}\;\cos \,2\theta } \right)\;d\theta } $$ $$I \; = \; \frac{1}{a}\int {d\theta } \;\; + \;\; \frac{1}{2}\int {{u^{ - 2}}\;du} \;\; + \;\; \frac{3}{{2{a^3}}}\int {\left( {1 + \cos \,2\theta } \right)\;d\theta } $$

Now all of the integrals are elementary forms, and we can take the antiderivatives en masse: $$I \; = \; \frac{1}{a}\theta \; + \; \frac{1}{2} \cdot \frac{{{u^{ - 1}}}}{{ - 1}} \; + \; \frac{3}{{2{a^3}}}\left( {\theta + \frac{{\sin \,2\theta }}{2}} \right) \; + \; C$$ Simplifying this (including the application of the double-angle formula $\sin \,2\theta = 2\,\sin \,\theta \,\cos \,\theta $) yields $$ I \; = \; \frac{1}{a}\theta \; - \; \frac{1}{{2u}} \; + \; \frac{3}{{2{a^3}}}\left( {\theta + \frac{{2\,\sin \,\theta \,\cos \,\theta }}{2}} \right) \; + \; C \; = \; \frac{1}{a}\theta \; - \; \frac{1}{{2u}} \; + \; \frac{3}{{2{a^3}}}\left( {\theta + \sin \,\theta \,\cos \,\theta } \right) \; + \; C $$

Now it is time to back-substitute these results with the original variable $x$. The first substitution is to change $\theta $ back to $v$, using $$\frac{v}{a} = \tan \;\theta \Rightarrow \theta = {\tan ^{ - 1}}\left( {\frac{v}{a}} \right) \Rightarrow \sin \;\theta = \frac{v}{{\sqrt {{v^2} + {a^2}} }}{\text{ and }}\cos \;\theta = \frac{a}{{\sqrt {{v^2} + {a^2}} }}$$ Therefore $$I = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{v}{a}} \right) - \frac{1}{{2u}} + \frac{3}{{2{a^3}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{v}{a}} \right) + \frac{v}{{\sqrt {{v^2} + {a^2}} }} \cdot \frac{a}{{\sqrt {{v^2} + {a^2}} }}} \right] + C$$ $$I = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{v}{a}} \right) - \frac{1}{{2u}} + \frac{3}{{2{a^3}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{v}{a}} \right) + \frac{{av}}{{{v^2} + {a^2}}}} \right] + C$$ We may then change $u$ and $v$ to $x$, remembering that $u = {x^2} - 4x + 6$, $v = x - 2$, and $a = \sqrt 2 $: $$I = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) - \frac{1}{{2\left( {{x^2} - 4x + 6} \right)}} + \frac{3}{{2{{\sqrt 2 }^3}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) + \frac{{\sqrt 2 \left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2} + 2}}} \right] + C$$

This is, in effect, the final answer. However, we can shorten it somewhat by distributing, simplifying like terms, and rationalizing some radicals (when it makes sense to do so). A few steps of algebra give us $$I = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) - \frac{1}{{2\left( {{x^2} - 4x + 6} \right)}} + \frac{3}{{4\sqrt 2 }}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) + \frac{{\sqrt 2 \left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2} + 2}}} \right] + C$$ $$I = \frac{4}{{4\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) - \frac{1}{{2\left( {{x^2} - 4x + 6} \right)}} + \frac{3}{{4\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) + \frac{3}{{4\sqrt 2 }} \cdot \frac{{\sqrt 2 \left( {x - 2} \right)}}{{\left( {{x^2} - 4x + 6} \right)}} + C$$ $$I = \frac{7}{{4\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) - \frac{2}{{4\left( {{x^2} - 4x + 6} \right)}} + \frac{{3\left( {x - 2} \right)}}{{4\left( {{x^2} - 4x + 6} \right)}} + C$$ and finally $$\boxed{\displaystyle{I = \frac{{7\sqrt 2 }}{8}{\tan ^{ - 1}}\left( {\frac{{x - 2}}{{\sqrt 2 }}} \right) + \frac{{3x - 8}}{{4\left( {{x^2} - 4x + 6} \right)}} + C}}$$ which agrees in all essential regards with the solution in the back of the textbook.