6.7 Integrals, exponential functions, and logarithms  (Page 4/4)

$y=\text{ln}\left(2x\right)$

$y=\text{ln}\left(2x+1\right)$

$y=\frac{1}{\text{ln}x}$

${\displaystyle \int \frac{dt}{3t}}$

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

[T] $y=\frac{\text{ln}\left(x\right)}{x}$

[T] $y=x\text{ln}\left(x\right)$

[T] $y={\text{log}}_{10}x$

[T] $y=\text{ln}\left(\text{sin}x\right)$

[T] $y=\text{ln}\left(\text{ln}x\right)$

[T] $y=7\text{ln}\left(4x\right)$

[T] $y=\text{ln}\left({\left(4x\right)}^7\right)$

[T] $y=\text{ln}\left(\text{tan}x\right)$

[T] $y=\text{ln}\left(\text{tan}\left(3x\right)\right)$

[T] $y=\text{ln}\left({\text{cos}}^{2}x\right)$

${\displaystyle {\int }_0^1\frac{dx}{3+x}}$

${\displaystyle {\int }_0^1\frac{dt}{3+2t}}$

${\displaystyle {\int }_0^2\frac{xdx}{x^2+1}}$

${\displaystyle {\int }_0^2\frac{x^{3}dx}{x^2+1}}$

${\displaystyle {\int }_2^e\frac{dx}{x\text{ln}x}}$

${\displaystyle {\int }_2^e\frac{dx}{{\left(x\text{ln}\left(x\right)\right)}^2}}$

${\displaystyle \int \frac{\text{cos}xdx}{\text{sin}x}}$

${\displaystyle {\int }_0^{\pi \text{/}4}\text{tan}xdx}$

${\displaystyle \int \text{cot}\left(3x\right)dx}$

${\displaystyle \int \frac{{\left(\text{ln}x\right)}^{2}dx}{x}}$

$y=\sqrt{x^2+1}$

$y=\sqrt{x^2+1}\sqrt{x^2-1}$

$y=e^{\text{sin}x}$

$y=x^{\mathrm{-1}\text{/}x}$

$y=e^{\left(ex\right)}$

$y=x^e$

$y=x^{\left(ex\right)}$

$y=\sqrt{x}\sqrt[3]{x}\sqrt[6]{x}$

$y=x^{\mathrm{-1}\text{/}\text{ln}x}$

$y=e^{\text{−}\text{ln}x}$

${\displaystyle {\int }_5^{10}\frac{dt}{t}-{\displaystyle {\int }_{5x}^{10x}\frac{dt}{t}}}$

${\displaystyle {\int }_1^{e^{\pi }}\frac{dx}{x}}+{\displaystyle {\int }_{\mathrm{-2}}^{\mathrm{-1}}\frac{dx}{x}}$

$\frac{d}{dx}{\displaystyle {\int }_x^1\frac{dt}{t}}$

$\frac{d}{dx}{\displaystyle {\int }_x^{x^2}\frac{dt}{t}}$

$\frac{d}{dx}\text{ln}\left(\text{sec}x+\text{tan}x\right)$

Find the area of the region enclosed by $x=1$ and $y=5$ above $y=\text{ln}x.$

[T] Find the arc length of $\text{ln}x$ from $x=1$ to $x=2.$

Find the area between $\text{ln}x$ and the x -axis from $x=1\text{to}x=2.$

Find the volume of the shape created when rotating this curve from $x=1\text{to}x=2$ around the x -axis, as pictured here.

[T] Find the surface area of the shape created when rotating the curve in the previous exercise from $x=1$ to $x=2$ around the x -axis.

Find the area of the hyperbolic quarter-circle enclosed by $x=2\text{and}y=2$ above $y=1\text{/}x.$

[T] Find the arc length of $y=1\text{/}x$ from $x=1\text{to}x=4.$

Find the area under $y=1\text{/}x$ and above the x -axis from $x=1\text{to}x=4.$

$\frac{d}{dx}\text{ln}\left(x+\sqrt{x^2+1}\right)=\frac{1}{\sqrt{1+x^2}}$

$\frac{d}{dx}\text{ln}\left(\frac{x-a}{x+a}\right)=\frac{2a}{\left(x^2-a^2\right)}$

$\frac{d}{dx}\text{ln}\left(\frac{1+\sqrt{1-x^2}}{x}\right)=-\frac{1}{x\sqrt{1-x^2}}$

$\frac{d}{dx}\text{ln}\left(x+\sqrt{x^2-a^2}\right)=\frac{1}{\sqrt{x^2-a^2}}$

${\displaystyle \int \frac{dx}{x\text{ln}\left(x\right)\text{ln}\left(\text{ln}x\right)}}=\text{ln}\left(\text{ln}\left(\text{ln}x\right)\right)+C$