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Calculus with the inverse hyperbolic functions

Total questions: 9

Worksheet time: 17mins

Name
Class
Date
1.

Which statements below are correct?


More than 1 correct answer is possible.

a)

ddx(arsinh⁡x)=1x2−1\frac{d}{dx}\left(\operatorname{arsinh}x\right)=\frac{1}{\sqrt{x^2-1}}

b)

ddx(arsinh⁡x)=11+x2\frac{d}{dx}\left(\operatorname{arsinh}x\right)=\frac{1}{\sqrt{1+x^2}}

c)

ddx(arcosh⁡x)=1x2−1\frac{d}{dx}\left(\operatorname{arcosh}x\right)=\frac{1}{\sqrt{x^2-1}}

d)

ddx(artanh⁡x)=11−x2\frac{d}{dx}\left(\operatorname{artanh}x\right)=\frac{1}{1-x^2}

2.

Which statements below are correct?


More than 1 correct answer is possible.

a)

∫(1x2−a2)dx=arcosh⁡(xa)+C, x>a\int_{ }^{ }\left(\frac{1}{\sqrt{x^2-a^2}}\right)dx=\operatorname{arcosh}\left(\frac{x}{a}\right)+C,\ x>a

b)

∫1a2−x2dx=1aartanh⁡(xa)+C, ∣x∣<a\int_{ }^{ }\frac{1}{a^2-x^2}dx=\frac{1}{a}\operatorname{artanh}\left(\frac{x}{a}\right)+C,\ \left|x\right|<a

c)

∫(1x2+a2)dx=1aarsinh⁡(xa)+C\int_{ }^{ }\left(\frac{1}{\sqrt{x^2+a^2}}\right)dx=\frac{1}{a}\operatorname{arsinh}\left(\frac{x}{a}\right)+C

d)

∫(1x2+a2)dx=arsinh⁡(xa)+C\int_{ }^{ }\left(\frac{1}{\sqrt{x^2+a^2}}\right)dx=\operatorname{arsinh}\left(\frac{x}{a}\right)+C

3.

The derivative of

 arcosh⁡2x\operatorname{arcosh}2x  
is

a)

 24x2−1\frac{2}{\sqrt{4x^2-1}}  

b)

 14x2−1\frac{1}{\sqrt{4x^2-1}}  

c)

 2x2−1\frac{2}{\sqrt{x^2-1}}  

d)

 1x2−1\frac{1}{\sqrt{x^2-1}}  

4.

The derivative of

 artanh⁡(1x)\operatorname{artanh}\left(\frac{1}{x}\right)  
is

a)

 11−x2\frac{1}{1-x^2}  

b)

 1x2−1\frac{1}{x^2-1}  

c)

 1x2(1−x2)\frac{1}{x^2\left(1-x^2\right)}  

d)

 x2x2−1\frac{x^2}{x^2-1}  

5.

The integral of

 ∫116+25x2dx\int_{ }^{ }\frac{1}{\sqrt{16+25x^2}}dx  
is

a)

 14arsinh⁡(5x4)+C\frac{1}{4}\operatorname{arsinh}\left(\frac{5x}{4}\right)+C  

b)

 15arsinh⁡(5x4)+C\frac{1}{5}\operatorname{arsinh}\left(\frac{5x}{4}\right)+C  

c)

 14arcosh⁡(5x4)+C\frac{1}{4}\operatorname{arcosh}\left(\frac{5x}{4}\right)+C  

d)

 15arsinh⁡(5x4)+C\frac{1}{5}\operatorname{arsinh}\left(\frac{5x}{4}\right)+C  

6.

The value of 
 ∫1.53116x2−9dx\int_{1.5}^3\frac{1}{\sqrt{16x^2-9}}dx  
is

a)

 ln⁡(4+152+3)\ln\left(\frac{4+\sqrt{15}}{2+\sqrt{3}}\right)  

b)

 14ln⁡(4−152−3)\frac{1}{4}\ln\left(\frac{4-\sqrt{15}}{2-\sqrt{3}}\right)  

c)

 14ln⁡(4+152+3)\frac{1}{4}\ln\left(\frac{4+\sqrt{15}}{2+\sqrt{3}}\right)  

d)

 ln⁡(4−152−3)\ln\left(\frac{4-\sqrt{15}}{2-\sqrt{3}}\right)  

7.

The integral ∫19+8x+2x2dx\int_{ }^{ }\frac{1}{\sqrt{9+8x+2x^2}}dx  

is

a)

 12arsinh⁡2(x+2)+C\frac{1}{\sqrt{2}}\operatorname{arsinh}\sqrt{2}\left(x+2\right)+C  

b)

 12arcosh⁡2(x+2)+C\frac{1}{\sqrt{2}}\operatorname{arcosh}\sqrt{2}\left(x+2\right)+C  

c)

 12arsinh⁡2(x+2)+C\frac{1}{2}\operatorname{arsinh}2\left(x+2\right)+C  

d)

 12arcosh⁡2(x+2)+C\frac{1}{2}\operatorname{arcosh}2\left(x+2\right)+C  

8.

The integral ∫13x2−12x−4dx\int_{ }^{ }\frac{1}{\sqrt{3x^2-12x-4}}dx  

is

a)

 arcosh⁡(3(x−2)4)+C\operatorname{arcosh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C  

b)

 13arcosh⁡(3(x−2)4)+C\frac{1}{\sqrt{3}}\operatorname{arcosh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C  

c)

 arsinh⁡(3(x−2)4)+C\operatorname{arsinh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C  

d)

 13arsinh⁡(3(x−2)4)+C\frac{1}{\sqrt{3}}\operatorname{arsinh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C  

9.

The integral ∫17−12x−4x2dx\int_{ }^{ }\frac{1}{7-12x-4x^2}dx  

is

a)

 14artanh⁡(2x+32)+C\frac{1}{4}\operatorname{artanh}\left(\frac{2x+3}{2}\right)+C  

b)

 14artanh⁡(x+32)+C\frac{1}{4}\operatorname{artanh}\left(\frac{x+3}{2}\right)+C  

c)

 18artanh⁡(x+34)+C\frac{1}{8}\operatorname{artanh}\left(\frac{x+3}{4}\right)+C  

d)

 18artanh⁡(2x+34)+C\frac{1}{8}\operatorname{artanh}\left(\frac{2x+3}{4}\right)+C