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Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of $f(x)=\frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}$ 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
Since $\text{deg}\left({\left({x}^{2}+1\right)}^{2}\right)=4>3=\text{deg}({x}^{3}),$ we can proceed with partial fraction decomposition. Note that ${({x}^{2}+1)}^{2}$ is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get
Finding a common denominator and equating the numerators gives
Solving, we obtain $A=1,$ $B=0,$ $C=\mathrm{-1},$ and $D=0.$ Substituting back into the integral, we have
Set up the partial fraction decomposition for $\int \frac{{x}^{2}+3x+1}{(x+2){(x-3)}^{2}{({x}^{2}+4)}^{2}}dx.$
$\frac{{x}^{2}+3x+1}{(x+2){\left(x-3\right)}^{2}{({x}^{2}+4)}^{2}}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{C}{{\left(x-3\right)}^{2}}+\frac{Dx+E}{{x}^{2}+4}+\frac{Fx+G}{{({x}^{2}+4)}^{2}}$
Express the rational function as a sum or difference of two simpler rational expressions.
$\frac{1}{(x-3)(x-2)}$
$\frac{{x}^{2}+1}{x(x+1)(x+2)}$
$-\frac{2}{x+1}+\frac{5}{2(x+2)}+\frac{1}{2x}$
$\frac{1}{{x}^{3}-x}$
$\frac{3{x}^{2}}{{x}^{2}+1}$ ( Hint: Use long division first.)
$\frac{2{x}^{4}}{{x}^{2}-2x}$
$2{x}^{2}+4x+8+\frac{16}{x-2}$
$\frac{1}{(x-1)({x}^{2}+1)}$
$\frac{1}{{x}^{2}(x-1)}$
$-\frac{1}{{x}^{2}}-\frac{1}{x}+\frac{1}{x-1}$
$\frac{x}{{x}^{2}-4}$
$\frac{1}{x(x-1)(x-2)(x-3)}$
$-\frac{1}{2(x-2)}+\frac{1}{2(x-1)}-\frac{1}{6x}+\frac{1}{6(x-3)}$
$\frac{1}{{x}^{4}-1}=\frac{1}{(x+1)(x-1)\left({x}^{2}+1\right)}$
$\frac{3{x}^{2}}{{x}^{3}-1}=\frac{3{x}^{2}}{(x-1)({x}^{2}+x+1)}$
$\frac{1}{x-1}+\frac{2x+1}{{x}^{2}+x+1}$
$\frac{2x}{{(x+2)}^{2}}$
$\frac{3{x}^{4}+{x}^{3}+20{x}^{2}+3x+31}{(x+1){\left({x}^{2}+4\right)}^{2}}$
$\frac{2}{x+1}+\frac{x}{{x}^{2}+4}-\frac{1}{{\left({x}^{2}+4\right)}^{2}}$
Use the method of partial fractions to evaluate each of the following integrals.
$\int \frac{dx}{(x-3)(x-2)}$
$\int \frac{3x}{{x}^{2}+2x-8}dx$
$\text{\u2212}\text{ln}\left|2-x\right|+2\phantom{\rule{0.1em}{0ex}}\text{ln}\left|4+x\right|+C$
$\int \frac{dx}{{x}^{3}-x}$
$\int \frac{x}{{x}^{2}-4}dx$
$\frac{1}{2}\text{ln}\left|4-{x}^{2}\right|+C$
$\int \frac{dx}{x(x-1)(x-2)(x-3)}$
$\int \frac{2{x}^{2}+4x+22}{{x}^{2}+2x+10}dx$
$2\left(x+\frac{1}{3}\text{arctan}\left(\frac{1+x}{3}\right)\right)+C$
$\int \frac{dx}{{x}^{2}-5x+6}$
$\int \frac{2-x}{{x}^{2}+x}dx$
$2\phantom{\rule{0.1em}{0ex}}\text{ln}\left|x\right|-3\phantom{\rule{0.1em}{0ex}}\text{ln}\left|1+x\right|+C$
$\int \frac{2}{{x}^{2}-x-6}dx$
$\int \frac{dx}{{x}^{3}-2{x}^{2}-4x+8}$
$\frac{1}{16}\left(\text{\u2212}\phantom{\rule{0.2em}{0ex}}\frac{4}{\mathrm{-2}+x}-\text{ln}\left|\mathrm{-2}+x\right|+\text{ln}\left|2+x\right|\right)+C$
$\int \frac{dx}{{x}^{4}-10{x}^{2}+9}$
Evaluate the following integrals, which have irreducible quadratic factors.
$\int \frac{2}{(x-4)\left({x}^{2}+2x+6\right)}dx$
$\frac{1}{30}(\mathrm{-2}\sqrt{5}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left[\frac{1+x}{\sqrt{5}}\right]+2\phantom{\rule{0.1em}{0ex}}\text{ln}\left|\mathrm{-4}+x\right|-\text{ln}\left|6+2x+{x}^{2}\right|)+C$
$\int \frac{{x}^{2}}{{x}^{3}-{x}^{2}+4x-4}dx$
$\int \frac{{x}^{3}+6{x}^{2}+3x+6}{{x}^{3}+2{x}^{2}}dx$
$-\frac{3}{x}+4\phantom{\rule{0.1em}{0ex}}\text{ln}\left|x+2\right|+x+C$
$\int \frac{x}{(x-1){\left({x}^{2}+2x+2\right)}^{2}}dx$
Use the method of partial fractions to evaluate the following integrals.
$\int \frac{3x+4}{\left({x}^{2}+4\right)(3-x)}dx$
$\text{\u2212}\text{ln}\left|3-x\right|+\frac{1}{2}\text{ln}\left|{x}^{2}+4\right|+C$
$\int \frac{2}{{(x+2)}^{2}(2-x)}dx$
$\int \frac{3x+4}{{x}^{3}-2x-4}dx$ ( Hint: Use the rational root theorem.)
$\text{ln}\left|x-2\right|-\frac{1}{2}\text{ln}\left|{x}^{2}+2x+2\right|+C$
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.)
$\int \frac{{e}^{x}dx}{{e}^{2x}-{e}^{x}}dx$
$\text{\u2212}x+\text{ln}\left|1-{e}^{x}\right|+C$
$\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx}{1-{\text{cos}}^{2}x}$
$\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{{\text{cos}}^{2}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x-6}dx$
$\frac{1}{5}\text{ln}\left|\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x+3}{\text{cos}\phantom{\rule{0.1em}{0ex}}x-2}\right|+C$
$\int \frac{1-\sqrt{x}}{1+\sqrt{x}}dx$
$\int \frac{dt}{{\left({e}^{t}-{e}^{\text{\u2212}t}\right)}^{2}}$
$\frac{1}{2-2{e}^{2t}}+C$
$\int \frac{1+{e}^{x}}{1-{e}^{x}}dx$
$\int \frac{dx}{1+\sqrt{x+1}}$
$2\sqrt{1+x}-2\phantom{\rule{0.1em}{0ex}}\text{ln}\left|1+\sqrt{1+x}\right|+C$
$\int \frac{dx}{\sqrt{x}+\sqrt[4]{x}}$
$\int \frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{\text{sin}\phantom{\rule{0.1em}{0ex}}x(1-\text{sin}\phantom{\rule{0.1em}{0ex}}x)}dx$
$\text{ln}\left|\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1-\text{sin}\phantom{\rule{0.1em}{0ex}}x}\right|+C$
$\int \frac{{e}^{x}}{{\left({e}^{2x}-4\right)}^{2}}dx$
$\underset{1}{\overset{2}{\int}}\frac{1}{{x}^{2}\sqrt{4-{x}^{2}}}dx$
$\frac{\sqrt{3}}{4}$
$\int \frac{1}{2+{e}^{\text{\u2212}x}}dx$
$\int \frac{1}{1+{e}^{x}}dx$
$x-\text{ln}\left(1+{e}^{x}\right)+C$
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 $y=1\text{/}\sqrt{x(3-x)},$ $y=0,$ $x=1,$ 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 $v(t)=\frac{88{t}^{2}}{{t}^{2}+1}.$ Find the distance that the particle has traveled after $t=5$ sec.
Solve the initial-value problem for x as a function of t.
$\left({t}^{2}-7t+12\right)\frac{dx}{dt}=1,\left(t>4,x(5)=0\right)$
$x=\text{\u2212}\text{ln}\left|t-3\right|+\text{ln}\left|t-4\right|+\text{ln}\phantom{\rule{0.1em}{0ex}}2$
$(t+5)\frac{dx}{dt}={x}^{2}+1,t>\text{\u2212}5,x(1)=\text{tan}\phantom{\rule{0.1em}{0ex}}1$
$\left(2{t}^{3}-2{t}^{2}+t-1\right)\frac{dx}{dt}=3,x(2)=0$
$x=\text{ln}\left|t-1\right|-\sqrt{2}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left(\sqrt{2}t\right)-\frac{1}{2}\text{ln}\left({t}^{2}+\frac{1}{2}\right)+\sqrt{2}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left(2\sqrt{2}\right)+\frac{1}{2}\text{ln}\phantom{\rule{0.1em}{0ex}}4.5$
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 $\int \frac{dx}{{x}^{3}+1}.$
$\frac{\text{arctan}\left[\frac{\mathrm{-1}+2x}{\sqrt{3}}\right]}{\sqrt{3}}+\frac{1}{3}\text{ln}\left|1+x\right|-\frac{1}{6}\text{ln}\left|1-x+{x}^{2}\right|+C$
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 $\int \frac{\sqrt[3]{x-8}}{x}dx}.$
$3{(\mathrm{-8}+x)}^{1\text{/}3}$
$\mathrm{-2}\sqrt{3}\phantom{\rule{0.1em}{0ex}}\text{arctan}\left[\frac{\mathrm{-1}+{(\mathrm{-8}+x)}^{1\text{/}3}}{\sqrt{3}}\right]$
$\mathrm{-2}\phantom{\rule{0.1em}{0ex}}\text{ln}\left[2+{(\mathrm{-8}+x)}^{1\text{/}3}\right]$
$+\text{ln}\left[4-2{(\mathrm{-8}+x)}^{1\text{/}3}+{(\mathrm{-8}+x)}^{2\text{/}3}\right]+C$
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