- #1

- 8

- 0

Find the following indefinite integral:

[tex] \int x.arsinh (x^2) dx [/tex]

Thanks very much for any help, its much appreciated.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Electrifying
- Start date

In summary: Thanks for the input!In summary, this conversation is about the integrals arcsinh and arcosh. The first step is to u-substitute and then integrate by parts. The u-sub is simple, so I'll skip that part. The second step is to use the wrong inspection method and get the wrong answer. Thanks for the help everyone.

- #1

- 8

- 0

Find the following indefinite integral:

[tex] \int x.arsinh (x^2) dx [/tex]

Thanks very much for any help, its much appreciated.

Physics news on Phys.org

- #2

Gold Member

- 1,222

- 22

- #3

- 8

- 0

Thanks for that, I got the following final answer:

[tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?

[tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?

Last edited:

- #4

- 33

- 0

I'm pretty sure that's wrong.

http://www.wolframalpha.com/input/?i=1/2(sinh^(-1)(x^2)-ln(1+x^4)^.5)'

http://www.wolframalpha.com/input/?i=1/2(sinh^(-1)(x^2)-ln(1+x^4)^.5)'

- #5

Mentor

- 37,463

- 9,713

The substitution should be u = arElectrifying said:

[tex] \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C [/tex]

Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?

- #6

Gold Member

- 1,222

- 22

Woops! I thought that read arcsin(x)! My bad.

- #7

- 8

- 0

Stimpon said:I'm pretty sure that's wrong.

http://www.wolframalpha.com/input/?i=1/2(sinh^(-1)(x^2)-ln(1+x^4)^.5)'

Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been taught to write inverse hyperbolic trig functions as arsinh, arcosh etc. rather than arc. Maybe its just an English convention, or did my sixth form teach me wrong?

- #8

Mentor

- 37,463

- 9,713

Did your teacher also write arcsin and arccos as arsin and arcos?Electrifying said:Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been taught to write inverse hyperbolic trig functions as arsinh, arcosh etc. rather than arc. Maybe its just an English convention, or did my sixth form teach me wrong?

- #9

- 8

- 0

- #10

Gold Member

- 1,222

- 22

- #11

Mentor

- 37,463

- 9,713

Well, you learn something every day.Char. Limit said:

A basic indefinite integral is the inverse operation of differentiation. It is a mathematical concept that involves finding the most general antiderivative of a given function.

A basic indefinite integral is represented using the integral symbol (∫) followed by the function to be integrated and the variable of integration.

The process for solving a basic indefinite integral involves applying the power rule, sum rule, and constant multiple rule to the given function. The resulting antiderivative is then expressed as an indefinite integral.

A basic indefinite integral does not have upper and lower limits, while a definite integral has specific values for the upper and lower limits. This means a basic indefinite integral represents a family of functions, while a definite integral represents a single value.

Basic indefinite integrals are important in mathematics because they help us find the area under the curve of a given function, which has numerous real-world applications in fields such as physics, engineering, and economics. They also provide a fundamental understanding of the relationship between a function and its derivative.

Share:

- Replies
- 3

- Views
- 497

- Replies
- 9

- Views
- 603

- Replies
- 3

- Views
- 768

- Replies
- 22

- Views
- 1K

- Replies
- 2

- Views
- 716

- Replies
- 19

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 3

- Views
- 939

- Replies
- 3

- Views
- 1K