${{\displaystyle \int }}^{\text{}}x{e}^{2x}dx=\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C$

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

$\text{−}{x}^2\text{cos}x+2x\text{sin}x+2\text{cos}x+C$

$\frac{\pi }{2}-1$

$\frac{1}{5}{\text{sin}}^{5}x+C$

$\frac{1}{3}{\text{sin}}^{3}x-\frac{1}{5}{\text{sin}}^{5}x+C$

$\frac{1}{2}x+\frac{1}{4}\text{sin}\left(2x\right)+C$

$\text{sin}x-\frac{1}{3}{\text{sin}}^{3}x+C$

$\frac{1}{2}x+\frac{1}{12}\text{sin}\left(6x\right)+C$

$\frac{1}{2}\text{sin}x+\frac{1}{22}\text{sin}\left(11x\right)+C$

$\frac{1}{6}{\text{tan}}^{6}x+C$

$\frac{1}{9}{\text{sec}}^{9}x-\frac{1}{7}{\text{sec}}^{7}x+C$

${\displaystyle \int }{\text{sec}}^{5}xdx=\frac{1}{4}{\text{sec}}^{3}x\text{tan}x+\frac{3}{4}{\displaystyle \int }{\text{sec}}^{3}x$

${{\displaystyle \int }}^{\text{}}125{\text{sin}}^3\theta d\theta$

${{\displaystyle \int }}^{\text{}}32{\text{tan}}^3\theta {\text{sec}}^3\theta d\theta$

$\text{ln}\left|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right|+C$

$x-5\text{ln}\left|x+2\right|+C$

$\frac{2}{5}\text{ln}\left|x+3\right|+\frac{3}{5}\text{ln}\left|x-2\right|+C$

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

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

Possible solutions include ${\text{sinh}}^{\mathrm{-1}}\left(\frac{x}{2}\right)+C$ and $\text{ln}\left|\sqrt{x^2+4}+x\right|+C.$

$\frac{24}{35}$

$\frac{17}{24}$

0.0074, 1.1%

$\frac{1}{192}$

$\frac{25}{36}$

$e^3,$ converges

$+\infty ,$ diverges

Since ${\displaystyle {\int }_e^{+\infty }\frac{1}{x}dx=\text{+}\infty },$ ${\displaystyle {\int }_e^{+\infty }\frac{\text{ln}x}{x}dx}$ diverges.

$u=x^3$

$u=y^3$

$u=\text{sin}(2x)$

$\text{−}x+x\text{ln}x+C$

$x{\text{tan}}^{\mathrm{-1}}x-\frac{1}{2}\text{ln}(1+x^2)+C$

$-\frac{1}{2}x\text{cos}\left(2x\right)+\frac{1}{4}\text{sin}\left(2x\right)+C$

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

$2x\text{cos}x+\left(\mathrm{-2}+x^2\right)\text{sin}x+C$

$\frac{1}{2}\left(1+2x\right)\left(\mathrm{-1}+\text{ln}(1+2x)\right)+C$

$\frac{1}{2}{e}^x\left(\text{−}\text{cos}x+\text{sin}x\right)+C$

$-\frac{e^{\text{−}{x}^2}}{2}+C$

$-\frac{1}{2}x\text{cos}\left[\text{ln}(2x)\right]+\frac{1}{2}x\text{sin}\left[\text{ln}(2x)\right]+C$

$2x-2x\text{ln}x+x{(\text{ln}x)}^2+C$

$\left(\text{−}\frac{x^3}{9}+\frac{1}{3}{x}^3\text{ln}x\right)+C$

$-\frac{1}{2}\sqrt{1-4x^2}+x{\text{cos}}^{\mathrm{-1}}(2x)+C$

$\text{−}\left(\mathrm{-2}+x^2\right)\text{cos}x+2x\text{sin}x+C$

$\text{−}x\left(\mathrm{-6}+x^2\right)\text{cos}x+3\left(\mathrm{-2}+x^2\right)\text{sin}x+C$

$\frac{1}{2}x\left(\text{−}\sqrt{1-\frac{1}{x^2}}+x·{\text{sec}}^{\mathrm{-1}}x\right)+C$

$\text{−}\text{cosh}x+x\text{sinh}x+C$

$\frac{1}{4}-\frac{3}{4{\text{e}}^2}$

2

$2\pi$

$\mathrm{-2}+\pi$

$\text{−}\text{sin}(x)+\text{ln}\left[\text{sin}(x)\right]\text{sin}x+C$

Answers vary

a. $\frac{2}{5}\left(1+x\right){\left(\mathrm{-3}+2x\right)}^{3\text{/}2}+C$ b. $\frac{2}{5}\left(1+x\right){\left(\mathrm{-3}+2x\right)}^{3\text{/}2}+C$

Do not use integration by parts. Choose u to be $\text{ln}x,$ and the integral is of the form ${\displaystyle \int u^{2}du}.$

Do not use integration by parts. Let $u=x^2-3,$ and the integral can be put into the form ${\displaystyle \int e^{u}du}.$

Do not use integration by parts. Choose u to be $u=3x^3+2$ and the integral can be put into the form ${\displaystyle \int \text{sin}(u)du}.$

The area under graph is 0.39535.

$2\pi e$

2.05

$12\pi$

$8{\pi }^2$

${\text{cos}}^{2}x$

$\frac{1-\text{cos}(2x)}{2}$

$\frac{{\text{sin}}^{4}x}{4}+C$

$\frac{1}{12}{\text{tan}}^6(2x)+C$

${\displaystyle \left(\frac{x}{2}\right)\left(\frac{x}{2}\right)+C_1 {\text{or sec}}^2\left(\frac{x}{2}\right)+C_2}$

$–\text{cos}x+\frac{1}{3}{\text{cos}}^{3}x+C$

$-\frac{1}{2}{\text{cos}}^{2}x+C$ or $\frac{1}{2}{\text{sin}}^{2}x+C$

$–\frac{1}{3}{\text{cos}}^{3}x+\frac{2}{5}{\text{cos}}^{5}x–\frac{1}{7}{\text{cos}}^{7}x+C$

$\frac{2}{3}{(\text{sin}x)}^{\frac{3}{2}}+C$

$\text{sec}x+C$

$\frac{1}{2}\text{sec}x\text{tan}x-\frac{1}{2}\text{ln}\left(\text{sec}x+\text{tan}x\right)+C$

$\frac{2\text{tan}x}{3}+\frac{1}{3}\text{sec}{(x)}^2\text{tan}x$ $=\text{tan}x+\frac{{\text{tan}}^{3}x}{3}+C$

$\text{−}\text{ln}\left|\text{cot}x+\text{csc}x\right|+C$

$\frac{{\text{sin}}^3\left(ax\right)}{3a}+C$

$\frac{\pi }{2}$

$\frac{x}{2}+\frac{1}{12}\text{sin}(6x)+C$

$x+C$

0

0

0

Approximately 0.239

$\sqrt{2}$

1.0

0

$\frac{3\theta }{8}-\frac{1}{4\pi }\text{sin}(2\pi \theta )+\frac{1}{32\pi }\text{sin}(4\pi \theta )+C=f\left(x\right)$

$\text{ln}\left(\sqrt{3}\right)$

${\displaystyle {\int }_{\text{−}\pi }^{\pi }\text{sin}(2x)\text{cos}(3x)}dx=0$

$\sqrt{\text{tan}(x)}x(\frac{8\text{tan}x}{21}+\frac{2}{7}\text{sec}{x}^2\text{tan}x)+C=f(x)$

The second integral is more difficult because the first integral is simply a u-substitution type.

$9{\text{tan}}^2\theta$

$a^2{\text{cosh}}^2\theta$

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

$\text{−}{(x+1)}^2+5$

$\text{ln}\left|\frac{\sqrt{x^2-a^2}+x}{a}\right|+C$

$\frac{1}{3}\text{ln}\left|\sqrt{9x^2+1}+3x\right|+C$

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

$9\left[\frac{x\sqrt{x^2+9}}{18}+\frac{1}{2}ln\left|\frac{\sqrt{x^2+9}}{3}+\frac{x}{3}\right|\right]+C$

$-\frac{1}{3}\sqrt{9-{\theta }^2}\left(18+{\theta }^2\right)+C$

$\frac{\left(\mathrm{-1}+x^2\right)\left(2+3x^2\right)\sqrt{x^6-x^8}}{15x^3}+C$

$-\frac{x}{9\sqrt{\mathrm{-9}+x^2}}+C$

$\frac{1}{2}\left(\text{ln}\left|x+\sqrt{x^2-1}\right|+x\sqrt{x^2-1}\right)+C$

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

$\frac{1}{8}\left(x\left(5-2x^2\right)\sqrt{1-x^2}+3\text{arcsin}x\right)+C$

$\text{ln}x-\text{ln}\left|1+\sqrt{1-x^2}\right|+C$

$-\frac{\sqrt{\mathrm{-1}+x^2}}{x}+\text{ln}\left|x+\sqrt{\mathrm{-1}+x^2}\right|+C$

$-\frac{\sqrt{1+x^2}}{x}+\text{arcsinh}x+C$

$-\frac{1}{1+x}+C$

$\text{arcsin}\left(\frac{\mathrm{x}-5}{5}\right)\text{+}C$

$\frac{9\pi }{2};$ area of a semicircle with radius 3

$\text{arcsin}(x)+C$ is the common answer.

$\frac{1}{2}\text{ln}\left(1+x^2\right)+C$ is the result using either method.

Use trigonometric substitution. Let $x=\text{sec}(\theta ).$

4.367

$\frac{{\pi }^2}{8}+\frac{\pi }{4}$

$y=\frac{1}{16}\text{ln}\left|\frac{x+8}{x-8}\right|+3$

24.6 m³

$\frac{2\pi }{3}$

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

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

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

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

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

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

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

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

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

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

$2\text{ln}\left|x\right|-3\text{ln}\left|1+x\right|+C$

$\frac{1}{16}\left(\text{−}\frac{4}{\mathrm{-2}+x}-\text{ln}\left|\mathrm{-2}+x\right|+\text{ln}\left|2+x\right|\right)+C$

$\frac{1}{30}(\mathrm{-2}\sqrt{5}\text{arctan}\left[\frac{1+x}{\sqrt{5}}\right]+2\text{ln}\left|\mathrm{-4}+x\right|-\text{ln}\left|6+2x+x^2\right|)+C$

$-\frac{3}{x}+4\text{ln}\left|x+2\right|+x+C$

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

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

$\text{−}x+\text{ln}\left|1-e^x\right|+C$

$\frac{1}{5}\text{ln}\left|\frac{\text{cos}x+3}{\text{cos}x-2}\right|+C$

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

$2\sqrt{1+x}-2\text{ln}\left|1+\sqrt{1+x}\right|+C$

$\text{ln}\left|\frac{\text{sin}x}{1-\text{sin}x}\right|+C$

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

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

$6x^{1\text{/}6}-3x^{1\text{/}3}+2\sqrt{x}-6\text{ln}\left(1+x^{1\text{/}6}\right)+C$

$\frac{4}{3}\pi \text{arctanh}\left[\frac{1}{3}\right]=\frac{1}{3}\pi \text{ln}4$

$x=\text{−}\text{ln}\left|t-3\right|+\text{ln}\left|t-4\right|+\text{ln}2$

$x=\text{ln}\left|t-1\right|-\sqrt{2}\text{arctan}\left(\sqrt{2}t\right)-\frac{1}{2}\text{ln}\left(t^2+\frac{1}{2}\right)+\sqrt{2}\text{arctan}\left(2\sqrt{2}\right)+\frac{1}{2}\text{ln}4.5$

$\frac{2}{5}\pi \text{ln}\frac{28}{13}$

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

2.0 in.²

$3{(\mathrm{-8}+x)}^{1\text{/}3}$
$\mathrm{-2}\sqrt{3}\text{arctan}\left[\frac{\mathrm{-1}+{(\mathrm{-8}+x)}^{1\text{/}3}}{\sqrt{3}}\right]$
$\mathrm{-2}\text{ln}\left[2+{(\mathrm{-8}+x)}^{1\text{/}3}\right]$
$+\text{ln}\left[4-2{(\mathrm{-8}+x)}^{1\text{/}3}+{(\mathrm{-8}+x)}^{2\text{/}3}\right]+C$

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

${\text{cosh}}^{\mathrm{-1}}\left(\frac{x+3}{3}\right)+C$

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

$\text{arcsin}\left(\frac{y}{2}\right)+C$

$-\frac{1}{2}\text{csc}\left(2w\right)+C$

$9-6\sqrt{2}$

$2-\frac{\pi }{2}$

$\frac{1}{12}{\text{tan}}^4\left(3x\right)-\frac{1}{6}{\text{tan}}^2\left(3x\right)+\frac{1}{3}\text{ln}\left|\text{sec}(3x)\right|+C$

$2\text{cot}\left(\frac{w}{2}\right)-2\text{csc}\left(\frac{w}{2}\right)+w+C$

$\frac{1}{5}\text{ln}\left|\frac{2(5+4\text{sin}t-3\text{cos}t)}{4\text{cos}t+3\text{sin}t}\right|$

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

$\text{−}{x}^3\text{cos}x+3x^2\text{sin}x+6x\text{cos}x-6\text{sin}x+C$

$\frac{1}{2}\left(x^2+\text{ln}\left|1+e^{\text{−}{x}^2}\right|\right)+C$