3.4 Partial fractions  (Page 4/5)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle {\int }_0^1\frac{e^x}{36-e^{2x}}dx}$ (Give the exact answer and the decimal equivalent. Round to five decimal places.)

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

${\displaystyle \int \frac{\text{sin}xdx}{1-{\text{cos}}^{2}x}}$

${\displaystyle \int \frac{\text{sin}x}{{\text{cos}}^{2}x+\text{cos}x-6}dx}$

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

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

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

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

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

${\displaystyle \int \frac{\text{cos}x}{\text{sin}x(1-\text{sin}x)}dx}$

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

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

${\displaystyle \int \frac{1}{2+e^{\text{−}x}}dx}$

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

${\displaystyle \int \frac{1}{t-\sqrt[3]{t}}dt}t=x^3$

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

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.

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

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

$(t+5)\frac{dx}{dt}=x^2+1,t>\text{−}5,x(1)=\text{tan}1$

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

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

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

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

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

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

${\displaystyle \int \frac{dx}{3-5\text{sin}x}}$

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

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

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