17.7 Rational Functions

In this section we present a few rules for finding antiderivatives of simple rational functions.

To antidifferentiate $${ {{P(x)}\over {(x-c)^n}}}$$ where $P$ is a polynomial, make the substitution $u=x-c$.

17.56   Example. To find $${ \int {{x^2+1}\over {(x-2)^2}}dx}$$.

Let $u=x-2$. Then $x=2+u$ so $dx=du$, and

$$\begin{eqnarray*} \int {{(x^2+1)}\over {(x-2)^2}}dx &=& \int {{(2+u)^2+1}\over {... ...-{5\over u}\\ &=& (x-2)+4\ln (\vert x-2\vert)-{5\over {(x-2)}}. \end{eqnarray*}$$

To find $${ \int {{R(x)}\over {(x-a)(x-b)}}dx}$$ where $a\neq b$ and $R$ is a polynomial of degree less than $2$.

We will find numbers $A$ and $B$ such that

$${{R(x)}\over {(x-a)(x-b)}}={A\over {(x-a)}}+{B\over {(x-b)}}.$$ (17.57)

Suppose (17.57) were valid. If we multiply both sides by $(x-a)$ we get

$${{R(x)}\over {(x-b)}}=A+{{B(x-a)}\over {x-b}}.$$

Now take the limit as $x$ goes to $a$ to get

$${{R(a)}\over {a-b}}=A.$$

The reason I took a limit here, instead of saying ``now for $x = a$ we get $\cdots$'' is that $a$ is not in the domain of the function we are considering. Similarly

$${{R(x)}\over {x-a}}={{A(x-b)}\over {x-a}}+B,$$

and if we take the limit as $x$ goes to $b$, we get

$${{R(b)}\over {b-a}}=B.$$

Thus,

$${{R(x)}\over {(x-a)(x-b)}}={1\over {a-b}}\Bigg[ {{R(a)}\over {x-a}}-{{R(b)}\over {x-b}}\Bigg].$$ (17.58)

I have now shown that if there are numbers $A$ and $B$ such that (17.57) holds, then (17.58) holds. Since I have not shown that such numbers exist, I will verify directly that (17.58) is valid. Write $R(x)=px+q$. Then

$$\begin{eqnarray*} {1\over {a-b}}\Bigg[ {{R(a)}\over {x-a}}-{{R(b)}\over {x-b}}\B... ...(x-a)(x-b)}} = {{R(x)}\over {(x-a)(x-b)}}.\mbox{ $\diamondsuit$} \end{eqnarray*}$$

17.59   Example. To find $${ \int {{x+1}\over {(x+2)(x+3)}}dx}$$.

Let $${ {{x+1}\over {(x+2)(x+3)}}={A\over {x+2}}+{B\over {x+3}}}$$.

Then

$${{x+1}\over {x+3}}=A+{{x+2}\over {x+3}}B,$$

so

$$A= {{-2+1}\over {-2+3}}=-1,$$

and

$${{x+1}\over {x+2}}=A{{(x+3)}\over{x+2}}+B,$$

so

$$B={{-3+1}\over {-3+2}}=2.$$

Hence

$$\begin{eqnarray*} \int {{x+1}\over {(x+2)(x+3)}}dx &=& \int {{-1}\over {x+2}}+{2\over {x+3}}dx \\ &=& -\ln (\vert x+2\vert)+2\ln (\vert x+3\vert). \end{eqnarray*}$$

In this example I did not use formula (17.58), because I find it easier to remember the procedure than the general formula. I do not need to check my answer, because my proof of (17.58) shows that the procedure always works. (In practice, I usually do check the result because I am likely to make an arithmetic error.)

To find $${ \int {{R(x)}\over {x^2+ax+b}}dx}$$ where $R$ is a polynomial of degree $<2$, and $x^2+ax+b$ does not factor as a product of two first degree polynomials.

Complete the square to write

$$x^2+ax+b=(x-m)^2+k.$$

Then $k>0$, since if $k=0$ then we have factored $x^2+ax+b$, and if $k < 0$ we can write $k=-n^2$, and then

$$(x-m)^2+k=(x-m)^2-n^2=\Big((x-m)-n\Big)\Big((x-m)+n\Big)$$

and again we get a factorization of $x^2+ax+b$. Since $k>0$, we can write $k=q^2$ for some $q\in\mbox{{\bf R}}$, and

$$x^2+ax+b=(x-m)^2+q^2.$$

Now

$$\int {{R(x)}\over {x^2+ax+b}}dx=\int {{R(x)}\over {(x-m)^2+q^2}}dx.$$

Make the substitution $u=x-m$ to get an antiderivative of the form

$$\begin{eqnarray*} \int {{Au+B}\over {u^2+q^2}}du &=& {A\over 2}\int {{2u}\over {... ...}}du \\ &=& {A\over 2}\ln (u^2+q^2)+B\int {1\over {u^2+q^2}}du. \end{eqnarray*}$$

The last antiderivative can be found by a trigonometric substitution.

17.60   Example. To find $${\int {u\over {4u^2+8u+7}}du}$$:

Let

$$\begin{eqnarray*} I=\int {u\over {4u^2+8u+7}}du &=& {1\over 4}\int {u\over {u^2+... ...r 4}}}du \\ &=& {1\over 4}\int {u\over {(u+1)^2+{3\over 4}}}du. \end{eqnarray*}$$

Let $t=u+1$, so $u=t-1$ and $du=dt$. Then

$$\begin{eqnarray*} I &=& {1\over 4}\int {{t-1}\over {t^2+{3\over 4}}}dt = {1\over... ... (u^2+2u+{7\over 4})-{1\over 4}\int {1\over {t^2+{3\over 4}}}dt. \end{eqnarray*}$$

Now let $${ t={{\sqrt 3}\over 2}\tan\theta}$$, so $${ dt={{\sqrt 3}\over 2}\sec^2\theta \; d\theta}$$, and $${ t^2+{3\over 4}={3\over 4}\sec^2\theta}$$. Then

$$\begin{eqnarray*} \int {1\over {t^2+{3\over 4}}}dt &=& \int {{ {{\sqrt 3}\over 2... ...&=& {2\over {\sqrt 3}}\arctan\Big( {{2u+2}\over {\sqrt 3}}\Big). \end{eqnarray*}$$

Hence,

$$I={1\over 8}\ln\Big( u^2+2u+{7\over 4}\Big)-{1\over {2\sqrt 3}}\arctan \Big( {{2u+2}\over {\sqrt 3}}\Big).$$

To find $${ \int {{R(x)}\over {x^2+ax+b}}dx}$$ where $R$ is a polynomial of degree $>1$.

First use long division to write

$${{R(x)}\over {x^2+ax+b}}=Q(x)+{{P(x)}\over {x^2+ax+b}}$$

where $Q$ is a polynomial, and $P$ is a polynomial of degree $\leq 1$. Then use one of the methods already discussed.

17.61   Example. To find $${\int {{x^3+1}\over {x^2+1}}dx}$$. By using long division, we get

$${{x^3+1}\over {x^2+1}}=x+{{-x+1}\over {x^2+1}}.$$

Hence

$$\begin{eqnarray*} \int {{x^3+1}\over {x^2+1}}dx &=& \int x - {x\over {x^2+1}}+{1... ...x^2+1}}dx &=& {1\over 2}x^2-{1\over 2}\ln (x^2+1)+\arctan (x). \end{eqnarray*}$$

17.62   Example. In exercise 17.7, you showed that $\ln\Big( \vert\sec(x)+\tan(x)\vert\Big)$ is an antiderivative for $\sec(x)$. The function $\ln\Big( \vert\sec(x)+\tan(x)\vert\Big)$ in that exercise appeared magically with no motivation. I will now derive the formula, using standard methods:

$$\int\sec(x)\;dx=\int {1\over {\cos(x)}}\;dx=\int{{\cos(x)}\over {\cos^2(x)}}\;dx=\int {{\cos(x)}\over {1-\sin^2(x)}}\;dx.$$

Now let $u=\sin(x)$. Then $du=\cos(x)\;dx$, and

$$\int\sec(x)\;dx=\int {{du}\over {1-u^2}}.$$

Suppose $${ {1\over {1-u^2}}={A\over {1-u}}+{B\over {1+u}}}$$. Then

$${1\over {1+u}}=A+{{B(1-u)}\over {(1+u)}},$$

and if we take the limit of both sides as $u\to 1$ we get $${A={1\over 2}}$$. Also

$${1\over {1-u}}={{A(1+u)}\over {1-u}}+B,$$

and if we take the limit as $u\to-1$, we get $${B={1\over 2}}$$. Thus

$$\begin{eqnarray*} \int\sec(x)\;dx &=& \int {1\over {1-u^2}}\;du \\ &=&{1\over 2... ...eft(\left\vert {{1+\sin(x)}\over {1-\sin(x)}}\right\vert\right). \end{eqnarray*}$$

Now

$${{1+\sin(x)}\over {1-\sin(x)}}={{1+\sin(x)}\over {1-\sin(x)}}... ...over {1-\sin^2(x)}}={{\Big(1+\sin(x)\Big)^2}\over {\cos^2(x)}},$$

so

$$\begin{eqnarray*} {1\over 2}\ln\left(\left\vert {{1+\sin(x)}\over {1-\sin(x)}}\r... ...right\vert\right)\\ &=&\ln\Big( \vert\sec(x)+\tan(x)\vert\Big), \end{eqnarray*}$$

and thus

$$\int\sec(x)\;dx=\ln\Big(\vert \sec(x)+\tan(x)\vert\Big).$$

17.63   Exercise. Criticize the following argument:

I want to find $${\int {{x^2}\over {x^2-1}}dx = \int {{x^2}\over {(x-1)(x+1)}}dx}$$. Suppose

$${{x^2}\over {(x-1)(x+1)}}={A\over {x-1}}+{B\over {x+1}}.$$

Then

$${{x^2}\over {x+1}}=A+{{(x-1)B}\over {x+1}}.$$

If we take the limit of both sides as $x\to 1$, we get $${ {1\over 2}=A}$$. Also

$${{x^2}\over {x-1}}={{A(x+1)}\over {x-1}}+B,$$

and if we take the limit of both sides as $x\to -1$, we get $${ -{1\over 2}=B}$$. Thus

$${{x^2}\over {x^2-1}}={1\over 2}{1\over {x-1}}-{1\over 2}{1\over {x+1}}.$$

Hence,

$$\int {{x^2}\over {x^2-1}}dx ={1\over 2}\ln (\vert x-1\vert)-{1\over 2}\ln (\vert x+1\vert).$$

17.64   Exercise. Find the following antiderivatives:

a)

$${ \int {1\over {4x^2-1}}dx}$$

b)

$${\int {1\over {4x^2+1}}dx}$$

c)

$${\int {{x+1}\over {x^2-6x+8}}dx}$$

d)

$${ \int {{x+1}\over {x^2-6x+9}}dx}$$

e)

$${\int {1\over {9x^2+6x+2}}dx}$$

f)

$${\int {{x^3}\over {x^2+1}}dx}$$

g)

$${\int {1\over {\sqrt{x^2+2x+2}}}dx}$$ A

17.65   Exercise. Find the following antiderivatives:

a)

$${\int{\cos(ax) \over \sin^3(ax)} dx }$$.

b)

$${\int{\sin(t)\cos(t) \over \cos^2(t) +1}dt}$$.

c)

$${\int{1\over(1-t)^3}dt}$$.

d)

$${\int{1 \over 5+4x+x^2} dx}$$.

e)

$${\int x^3 \sqrt{x^2+1} dx}$$.

f)

$${\int {1\over\sqrt{ -3 -4x -x^2}}}$$.

g)

$${ \int{\sin(2\theta) \over \cos^2(\theta) - \sin^2(\theta)} d\theta }$$.

h)

$${\int(1+\tan(u))^2du}$$.

i)

Choose a number $p$, and find $${\int x^p(x^{10} -2)^{10} dx}$$.

j)

Choose a number $q$, and find $${\int x^q e^{-{1\over x}}dx}$$.

k)

$${\int x e^{-x^2} dx}$$.

l)

$${\int{ u^3\over 1+u^2} du}$$.

m)

$${\int x^2 \arctan(x) dx}$$.

n)

$${ \int x^3(1+x)^{1\over 4} dx}$$.

o)

$${\int x e^{2x} dx}$$.

p)

$${\int \arcsin(x) dx}$$.