OBJECTIVE

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

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

NEED TO REVIEW?

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

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.

Find

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

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*}$$

(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}$$

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.

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.