- #1

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Find the following indefinite integral:

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

Thanks very much for any help, its much appreciated.

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- Thread starter Electrifying
- Start date

- #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.

- #2

Char. Limit

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- #3

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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

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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

Mark44

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The substitution should be u = arThanks 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?

- #6

Char. Limit

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

- #7

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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

Mark44

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Did your teacher also write arcsin and arccos as arsin and arcos?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

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- #10

Char. Limit

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- #11

Mark44

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Well, ya learn something every day.

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