# 3.4 Partial fractions  (Page 4/5)

 Page 4 / 5

## Finding a volume

Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of $f\left(x\right)=\frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}$ and the x -axis over the interval $\left[0,1\right]$ about the y -axis.

Let’s begin by sketching the region to be revolved (see [link] ). From the sketch, we see that the shell method is a good choice for solving this problem.

The volume is given by

$V=2\pi {\int }_{0}^{1}x·\frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}dx=2\pi {\int }_{0}^{1}\frac{{x}^{3}}{{\left({x}^{2}+1\right)}^{2}}dx.$

Since $\text{deg}\left({\left({x}^{2}+1\right)}^{2}\right)=4>3=\text{deg}\left({x}^{3}\right),$ we can proceed with partial fraction decomposition. Note that ${\left({x}^{2}+1\right)}^{2}$ is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get

$\frac{{x}^{3}}{{\left({x}^{2}+1\right)}^{2}}=\frac{Ax+B}{{x}^{2}+1}+\frac{Cx+D}{{\left({x}^{2}+1\right)}^{2}}.$

Finding a common denominator and equating the numerators gives

${x}^{3}=\left(Ax+B\right)\left({x}^{2}+1\right)+Cx+D.$

Solving, we obtain $A=1,$ $B=0,$ $C=-1,$ and $D=0.$ Substituting back into the integral, we have

$\begin{array}{cc}\hfill V& =2\pi \underset{0}{\overset{1}{\int }}\frac{{x}^{3}}{{\left({x}^{2}+1\right)}^{2}}dx\hfill \\ & =2\pi \underset{0}{\overset{1}{\int }}\left(\frac{x}{{x}^{2}+1}-\frac{x}{{\left({x}^{2}+1\right)}^{2}}\right)dx\hfill \\ & =2\pi \left(\frac{1}{2}\text{ln}\left({x}^{2}+1\right)+\frac{1}{2}·\frac{1}{{x}^{2}+1}\right)|{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}1\\ \end{array}}\hfill \\ & =\pi \left(\text{ln}\phantom{\rule{0.1em}{0ex}}2-\frac{1}{2}\right).\hfill \end{array}$

Set up the partial fraction decomposition for $\int \frac{{x}^{2}+3x+1}{\left(x+2\right){\left(x-3\right)}^{2}{\left({x}^{2}+4\right)}^{2}}dx.$

$\frac{{x}^{2}+3x+1}{\left(x+2\right){\left(x-3\right)}^{2}{\left({x}^{2}+4\right)}^{2}}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{C}{{\left(x-3\right)}^{2}}+\frac{Dx+E}{{x}^{2}+4}+\frac{Fx+G}{{\left({x}^{2}+4\right)}^{2}}$

## Key concepts

• Partial fraction decomposition is a technique used to break down a rational function into a sum of simple rational functions that can be integrated using previously learned techniques.
• When applying partial fraction decomposition, we must make sure that the degree of the numerator is less than the degree of the denominator. If not, we need to perform long division before attempting partial fraction decomposition.
• The form the decomposition takes depends on the type of factors in the denominator. The types of factors include nonrepeated linear factors, repeated linear factors, nonrepeated irreducible quadratic factors, and repeated irreducible quadratic factors.

Express the rational function as a sum or difference of two simpler rational expressions.

$\frac{1}{\left(x-3\right)\left(x-2\right)}$

$\frac{{x}^{2}+1}{x\left(x+1\right)\left(x+2\right)}$

$-\frac{2}{x+1}+\frac{5}{2\left(x+2\right)}+\frac{1}{2x}$

$\frac{1}{{x}^{3}-x}$

$\frac{3x+1}{{x}^{2}}$

$\frac{1}{{x}^{2}}+\frac{3}{x}$

$\frac{3{x}^{2}}{{x}^{2}+1}$ ( Hint: Use long division first.)

$\frac{2{x}^{4}}{{x}^{2}-2x}$

$2{x}^{2}+4x+8+\frac{16}{x-2}$

$\frac{1}{\left(x-1\right)\left({x}^{2}+1\right)}$

$\frac{1}{{x}^{2}\left(x-1\right)}$

$-\frac{1}{{x}^{2}}-\frac{1}{x}+\frac{1}{x-1}$

$\frac{x}{{x}^{2}-4}$

$\frac{1}{x\left(x-1\right)\left(x-2\right)\left(x-3\right)}$

$-\frac{1}{2\left(x-2\right)}+\frac{1}{2\left(x-1\right)}-\frac{1}{6x}+\frac{1}{6\left(x-3\right)}$

$\frac{1}{{x}^{4}-1}=\frac{1}{\left(x+1\right)\left(x-1\right)\left({x}^{2}+1\right)}$

$\frac{3{x}^{2}}{{x}^{3}-1}=\frac{3{x}^{2}}{\left(x-1\right)\left({x}^{2}+x+1\right)}$

$\frac{1}{x-1}+\frac{2x+1}{{x}^{2}+x+1}$

$\frac{2x}{{\left(x+2\right)}^{2}}$

$\frac{3{x}^{4}+{x}^{3}+20{x}^{2}+3x+31}{\left(x+1\right){\left({x}^{2}+4\right)}^{2}}$

$\frac{2}{x+1}+\frac{x}{{x}^{2}+4}-\frac{1}{{\left({x}^{2}+4\right)}^{2}}$

Use the method of partial fractions to evaluate each of the following integrals.

$\int \frac{dx}{\left(x-3\right)\left(x-2\right)}$

$\int \frac{3x}{{x}^{2}+2x-8}dx$

$\text{−}\text{ln}|2-x|+2\phantom{\rule{0.1em}{0ex}}\text{ln}|4+x|+C$

$\int \frac{dx}{{x}^{3}-x}$

$\int \frac{x}{{x}^{2}-4}dx$

$\frac{1}{2}\text{ln}|4-{x}^{2}|+C$

$\int \frac{dx}{x\left(x-1\right)\left(x-2\right)\left(x-3\right)}$

$\int \frac{2{x}^{2}+4x+22}{{x}^{2}+2x+10}dx$

$2\left(x+\frac{1}{3}\text{arctan}\left(\frac{1+x}{3}\right)\right)+C$

$\int \frac{dx}{{x}^{2}-5x+6}$

$\int \frac{2-x}{{x}^{2}+x}dx$

$2\phantom{\rule{0.1em}{0ex}}\text{ln}|x|-3\phantom{\rule{0.1em}{0ex}}\text{ln}|1+x|+C$

$\int \frac{2}{{x}^{2}-x-6}dx$

$\int \frac{dx}{{x}^{3}-2{x}^{2}-4x+8}$

$\frac{1}{16}\left(\text{−}\phantom{\rule{0.2em}{0ex}}\frac{4}{-2+x}-\text{ln}|-2+x|+\text{ln}|2+x|\right)+C$

$\int \frac{dx}{{x}^{4}-10{x}^{2}+9}$

Evaluate the following integrals, which have irreducible quadratic factors.

$\int \frac{2}{\left(x-4\right)\left({x}^{2}+2x+6\right)}dx$

$\frac{1}{30}\left(-2\sqrt{5}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left[\frac{1+x}{\sqrt{5}}\right]+2\phantom{\rule{0.1em}{0ex}}\text{ln}|-4+x|-\text{ln}|6+2x+{x}^{2}|\right)+C$

$\int \frac{{x}^{2}}{{x}^{3}-{x}^{2}+4x-4}dx$

$\int \frac{{x}^{3}+6{x}^{2}+3x+6}{{x}^{3}+2{x}^{2}}dx$

$-\frac{3}{x}+4\phantom{\rule{0.1em}{0ex}}\text{ln}|x+2|+x+C$

$\int \frac{x}{\left(x-1\right){\left({x}^{2}+2x+2\right)}^{2}}dx$

Use the method of partial fractions to evaluate the following integrals.

$\int \frac{3x+4}{\left({x}^{2}+4\right)\left(3-x\right)}dx$

$\text{−}\text{ln}|3-x|+\frac{1}{2}\text{ln}|{x}^{2}+4|+C$

$\int \frac{2}{{\left(x+2\right)}^{2}\left(2-x\right)}dx$

$\int \frac{3x+4}{{x}^{3}-2x-4}dx$ ( Hint: Use the rational root theorem.)

$\text{ln}|x-2|-\frac{1}{2}\text{ln}|{x}^{2}+2x+2|+C$

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.

${\int }_{0}^{1}\frac{{e}^{x}}{36-{e}^{2x}}dx$ (Give the exact answer and the decimal equivalent. Round to five decimal places.)

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

$\text{−}x+\text{ln}|1-{e}^{x}|+C$

$\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx}{1-{\text{cos}}^{2}x}$

$\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{{\text{cos}}^{2}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x-6}dx$

$\frac{1}{5}\text{ln}|\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x+3}{\text{cos}\phantom{\rule{0.1em}{0ex}}x-2}|+C$

$\int \frac{1-\sqrt{x}}{1+\sqrt{x}}dx$

$\int \frac{dt}{{\left({e}^{t}-{e}^{\text{−}t}\right)}^{2}}$

$\frac{1}{2-2{e}^{2t}}+C$

$\int \frac{1+{e}^{x}}{1-{e}^{x}}dx$

$\int \frac{dx}{1+\sqrt{x+1}}$

$2\sqrt{1+x}-2\phantom{\rule{0.1em}{0ex}}\text{ln}|1+\sqrt{1+x}|+C$

$\int \frac{dx}{\sqrt{x}+\sqrt[4]{x}}$

$\int \frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{\text{sin}\phantom{\rule{0.1em}{0ex}}x\left(1-\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)}dx$

$\text{ln}|\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1-\text{sin}\phantom{\rule{0.1em}{0ex}}x}|+C$

$\int \frac{{e}^{x}}{{\left({e}^{2x}-4\right)}^{2}}dx$

$\underset{1}{\overset{2}{\int }}\frac{1}{{x}^{2}\sqrt{4-{x}^{2}}}dx$

$\frac{\sqrt{3}}{4}$

$\int \frac{1}{2+{e}^{\text{−}x}}dx$

$\int \frac{1}{1+{e}^{x}}dx$

$x-\text{ln}\left(1+{e}^{x}\right)+C$

Use the given substitution to convert the integral to an integral of a rational function, then evaluate.

$\int \frac{1}{t-\sqrt[3]{t}}dt\phantom{\rule{0.1em}{0ex}}t={x}^{3}$

$\int \frac{1}{\sqrt{x}+\sqrt[3]{x}}dx;x={u}^{6}$

$6{x}^{1\text{/}6}-3{x}^{1\text{/}3}+2\sqrt{x}-6\phantom{\rule{0.1em}{0ex}}\text{ln}\left(1+{x}^{1\text{/}6}\right)+C$

Graph the curve $y=\frac{x}{1+x}$ over the interval $\left[0,5\right].$ Then, find the area of the region bounded by the curve, the x -axis, and the line $x=4.$

Find the volume of the solid generated when the region bounded by $y=1\text{/}\sqrt{x\left(3-x\right)},$ $y=0,$ $x=1,$ and $x=2$ is revolved about the x- axis.

$\frac{4}{3}\pi \phantom{\rule{0.1em}{0ex}}\text{arctanh}\left[\frac{1}{3}\right]=\frac{1}{3}\pi \phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}4$

The velocity of a particle moving along a line is a function of time given by $v\left(t\right)=\frac{88{t}^{2}}{{t}^{2}+1}.$ Find the distance that the particle has traveled after $t=5$ sec.

Solve the initial-value problem for x as a function of t.

$\left({t}^{2}-7t+12\right)\frac{dx}{dt}=1,\left(t>4,x\left(5\right)=0\right)$

$x=\text{−}\text{ln}|t-3|+\text{ln}|t-4|+\text{ln}\phantom{\rule{0.1em}{0ex}}2$

$\left(t+5\right)\frac{dx}{dt}={x}^{2}+1,t>\text{−}5,x\left(1\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}1$

$\left(2{t}^{3}-2{t}^{2}+t-1\right)\frac{dx}{dt}=3,x\left(2\right)=0$

$x=\text{ln}|t-1|-\sqrt{2}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left(\sqrt{2}t\right)-\frac{1}{2}\text{ln}\left({t}^{2}+\frac{1}{2}\right)+\sqrt{2}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left(2\sqrt{2}\right)+\frac{1}{2}\text{ln}\phantom{\rule{0.1em}{0ex}}4.5$

Find the x -coordinate of the centroid of the area bounded by

$y\left({x}^{2}-9\right)=1,$ $y=0,x=4,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}x=5.$ (Round the answer to two decimal places.)

Find the volume generated by revolving the area bounded by $y=\frac{1}{{x}^{3}+7{x}^{2}+6x}x=1,x=7,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=0$ about the y -axis.

$\frac{2}{5}\pi \phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}\frac{28}{13}$

Find the area bounded by $y=\frac{x-12}{{x}^{2}-8x-20},$ $y=0,x=2,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}x=4.$ (Round the answer to the nearest hundredth.)

Evaluate the integral $\int \frac{dx}{{x}^{3}+1}.$

$\frac{\text{arctan}\left[\frac{-1+2x}{\sqrt{3}}\right]}{\sqrt{3}}+\frac{1}{3}\text{ln}|1+x|-\frac{1}{6}\text{ln}|1-x+{x}^{2}|+C$

For the following problems, use the substitutions $\text{tan}\left(\frac{x}{2}\right)=t,$ $dx=\frac{2}{1+{t}^{2}}dt,$ $\text{sin}\phantom{\rule{0.1em}{0ex}}x=\frac{2t}{1+{t}^{2}},$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x=\frac{1-{t}^{2}}{1+{t}^{2}}.$

$\int \frac{dx}{3-5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x}$

Find the area under the curve $y=\frac{1}{1+\text{sin}\phantom{\rule{0.1em}{0ex}}x}$ between $x=0$ and $x=\pi .$ (Assume the dimensions are in inches.)

2.0 in. 2

Given $\text{tan}\left(\frac{x}{2}\right)=t,$ derive the formulas $dx=\frac{2}{1+{t}^{2}}dt,$ $\text{sin}\phantom{\rule{0.1em}{0ex}}x=\frac{2t}{1+{t}^{2}},$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x=\frac{1-{t}^{2}}{1+{t}^{2}}.$

Evaluate $\int \frac{\sqrt[3]{x-8}}{x}dx.$

$3{\left(-8+x\right)}^{1\text{/}3}$
$-2\sqrt{3}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left[\frac{-1+{\left(-8+x\right)}^{1\text{/}3}}{\sqrt{3}}\right]$
$-2\phantom{\rule{0.1em}{0ex}}\text{ln}\left[2+{\left(-8+x\right)}^{1\text{/}3}\right]$
$+\text{ln}\left[4-2{\left(-8+x\right)}^{1\text{/}3}+{\left(-8+x\right)}^{2\text{/}3}\right]+C$

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!