7.4 Substitution: Integrands Containing ax2+bx+c

496

OBJECTIVE

When you finish this section, you should be able to:

  1. Find an integral that contains a quadratic expression (p. 496)

1 Find an Integral That Contains a Quadratic Expression

NEED TO REVIEW?

Completing the square is discussed in Appendix A.1, pp. A-2 to A-3.

Integrals containing \(ax^{2}+bx+c\), where \(a \ne 0\), \(b\) and \(c\) are real numbers, can often be integrated by completing the square. We complete the square of \(ax^{2}+bx+c\) as follows: \[ \begin{array}{rcl@{\quad}l} ax^{2}+bx+c &=& a\left( x^{2}+\dfrac{b}{a}x\right) +c & {\color{#0066A7}{\hbox{Factor \(a\) from the first two terms.}}} \\ &=& a\left( x^{2}+\dfrac{b}{a}x+\dfrac{b^{2}}{4a^{2}}\right) +c-\dfrac{b^{2}}{4a} & {\color{#0066A7}{\hbox{Add and subtract \(a\) \(\left( \dfrac{1}{2} \cdot \dfrac{b}{a}\right) ^{\!\!\!2} =\dfrac{b^{2}}{4a}\).}}} \\ &=&a\left( x+\dfrac{b}{2a}\right) ^{2}+\left( c-\dfrac{b^{2}}{4a}\right) & {\color{#0066A7}{\hbox{Factor \(x^{2} + \dfrac{b}{a} x+ \dfrac{b^{2}}{4a^{2}} = \left(x+\dfrac{b}{2a}\right) ^{\!\!\!2}\).}}} \end{array} \]

After completing the square, we use the substitution \[ u=x+\frac{b}{2a} \]

to express the original quadratic expression \(ax^{2}+bx+c\) in the simpler form \(au^{2}+r\), where \(r=c- \dfrac{b^{2}}{4a}\).

Finding an Integral Containing \(x^{2}+6x+10\)

Find \(\int \dfrac{dx}{x^{2}+6x+10}\).

Solution The integrand contains the quadratic expression \(x^{2}+6x+10\). So, we complete the square. \[ x^{2}+6x+10=(x^{2}+6x+9)+1=(x+3)^{2}+1 \]

Now we write the integral as \[ \int \frac{dx}{x^{2}+6x+10}=\int \frac{dx}{(x+3)^{2}+1} \]

and use the substitution \(u=x+3\). Then \(du=dx\), and \[ \int \frac{dx}{x^{2}+6x+10}=\int \frac{dx}{(x+3)^{2}+1}=\int \frac{du}{ u^{2}+1}=\tan ^{-1}u+C=\tan ^{-1}(x+3)+C \]

RECALL

\(\int \dfrac{dx}{a^{2}+x^{2}} =\dfrac{1}{a}\tan ^{-1} \dfrac{x}{a} +C\!\)

NOW WORK

Problem 1.

497

Finding Integrals Containing \(x^{2}+x+1\)

Find

  1. \(\int \dfrac{dx}{x^{2}+x+1}\)
  2. \(\int \dfrac{x\,dx}{x^{2}+x+1}\)

Solution (a) The integrand contains the quadratic expression \(x^{2}+x+1\). So, we complete the square in the denominator. \[ \begin{eqnarray*} \int \dfrac{dx}{x^{2}+x+1}\underset{\underset{\color{#0066A7}{\hbox{Complete the square}}}{\color{#0066A7}{\uparrow}}} {=}\int \dfrac{dx}{\left( x^{2}+x+\dfrac{1}{4} \right) +\left( 1-\dfrac{1}{4}\right) }=\int \dfrac{dx}{\left( x+\dfrac{1}{2} \right) ^{2}+\dfrac{3}{4}}\\[-2.1pc] \end{eqnarray*} \]

RECALL

\(\int \dfrac{du}{u^2+a^2} = \dfrac{1}{a}\tan^{-1} \dfrac{u}{a} + C\)

Now we use the substitution \(u=x+\dfrac{1}{2}\). Then \(du=dx\). \[ \begin{eqnarray*} \int \dfrac{dx}{x^{2}+x+1}&=&\int \dfrac{dx}{\left( x+\dfrac{1}{2}\right) ^{2}+\dfrac{3}{4}}=\int \dfrac{du}{u^{2}+\dfrac{3}{4}} \qquad {\color{#0066A7}{\hbox{\(a=\dfrac{\sqrt{3}}{2}\)}}} \\ &=& \dfrac{1}{\dfrac{\sqrt{3}}{2}} \tan^{-1} \dfrac{u}{\dfrac{\sqrt{3}}{2}} + C = \dfrac{2}{\sqrt{3}}\tan^{-1} \dfrac{2u}{\sqrt{3}} +C\\ &=&\dfrac{2\sqrt{3}}{3}\tan ^{-1} \dfrac{2x+1}{\sqrt{3}} +C\qquad {\color{#0066A7}{\hbox{\(u=x+\dfrac{1}{2}\)}}} \end{eqnarray*} \]

NOTE

We could also complete the square and let \(u=x+\dfrac{1}{2}\).

(b) This problem requires some imagination. We force the derivative of the denominator to appear in the numerator by using the following algebraic manipulations: \[ \begin{array}{@{\hspace*{0pt}}rcl@{\qquad}l} \int \dfrac{x~dx}{x^{2}+x+1} &=&\dfrac{1}{2}\int \dfrac{2x~dx}{x^{2}+x+1} & {\color{#0066A7}{\hbox{Multiply the integrand by \(\dfrac{2}{2}.\)}}} \\ &=&\dfrac{1}{2}\int \dfrac{[ ( 2x+1) -1] dx}{x^{2}+x+1} & {\color{#0066A7}{\hbox{Add and subtract 1 in}}} \\[-6pt] &&&{\color{#0066A7}{\hbox{the numerator to get \(2x+1\).}}}\\ &=&\dfrac{1}{2}\int \dfrac{( 2x+1) dx}{x^{2}+x+1}-\dfrac{1}{2}\int \dfrac{dx}{x^{2}+x+1} & {\color{#0066A7}{\hbox{Write the integral as the}}}\\[-6pt] &&&{\color{#0066A7}{\hbox{sum of two integrals.}}} \end{array} \]

RECALL

\(\int \dfrac{g'(x)}{g(x)}\,dx = \ln |g(x)| +C \)

Now we find each integral separately. We found the integral on the right in (a). In the integral on the left, the numerator equals the derivative of the denominator. \[ \begin{eqnarray*} \frac{1}{2}\int \dfrac{( 2x+1) dx}{x^{2}+x+1}&=& \frac{1}{2}\ln \left\vert x^{2}+x+1\right\vert \end{eqnarray*} \]

So, \[ \begin{eqnarray*} \int \dfrac{x~dx}{x^{2}+x+1} &=&\dfrac{1}{2}\int \dfrac{\left( 2x+1\right) dx}{x^{2}+x+1}-\dfrac{1}{2}\int \dfrac{dx}{x^{2}+x+1}\\[5pt] &=&\dfrac{1}{2}\ln \left\vert x^{2}+x+1\right\vert -\dfrac{1}{2}\left[ \dfrac{2\sqrt{3}}{3}\tan ^{-1} \dfrac{2x+1}{\sqrt{3}} \right] +C \nonumber \\[5pt] &=&\dfrac{1}{2}\ln \left\vert x^{2}+x+1\right\vert -\dfrac{\sqrt{3}}{3}\tan ^{-1} \dfrac{2x+1}{\sqrt{3}} +C \end{eqnarray*} \]

NOW WORK

Problem 7.

498

Finding an Integral Containing \(2x-x^{2}\)

Find \(\int \dfrac{dx}{\sqrt{2x-x^{2}}}\).

Solution The integrand contains the quadratic expression \(2x-x^{2}\), so we complete the square. \[ \begin{eqnarray*} 2x-x^{2}&=&-x^{2}+2x=-(x^{2}-2x) =-(x^{2}-2x+1)+1\\ &=&-( x-1) ^{2}+1=1-(x-1)^{2} \end{eqnarray*} \]

Then \[ \begin{eqnarray*} \int\! \frac{dx}{\sqrt{2x-x^{2}}}&=&\!\int\! \frac{dx}{\sqrt{1-(x-1)^{2}}}&&\underset{\underset{\underset{\color{#0066A7}{\hbox{\(du=dx\)}}}{\color{#0066A7}{\hbox{\(u=x-1\)}}}}{\color{#0066A7}{\uparrow }}}{=}\!\int\! \frac{du}{\sqrt{ 1-u^{2}}}=\sin ^{-1}u+C=\sin ^{-1}(x-1)+C \\[-11pt] \end{eqnarray*} \]

RECALL

\(\int \dfrac{dx}{\sqrt{a^{2}-x^{2}}} = \sin ^{-1} \dfrac{x}{a} +C\)

NOW WORK

Problem 23.