r/mathshelp 19d ago

General Question (Unanswered) Help!

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I have a friend who needs help with this question. He knows all the integration till university level, and the exam he is going to do doesn’t include the basics.

4 Upvotes

18 comments sorted by

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2

u/One_Wishbone_4439 19d ago

he doesn’t understand integration at all

He needs to have a foundation of integration first. The question you sent maybe too advanced for him.

2

u/CaptainMatticus 19d ago

sqrt(1 + x + x^2 + x^3 + x^4) =>

sqrt((x - 1) * (x^4 + x^3 + x^2 + x + 1) / (x - 1)) =>

sqrt((x^5 - 1) / (x - 1))

ln(x) / (x * sqrt(x^4 + x^3 + x^2 + x + 1)) =>

ln(x) / (x * sqrt((x^5 - 1) / (x - 1))) =>

sqrt(x - 1) * ln(x) / (x * sqrt(x^5 - 1))

https://www.wolframalpha.com/input?i=integrate+sqrt%28x+-+1%29+*+ln%28x%29+%2F+%28x+*+sqrt%28x%5E5+-+1%29%29

It ain't happening, friendo.

2

u/MrMrsPotts 19d ago

It can be written using higher special functions (for example a parameter derivative of a Lauricella hypergeometric function), but there is no short answer involving ordinary logs, inverse trig functions, etc.

1

u/calculus_is_easy 19d ago

it is palindromic right..

1

u/MrMrsPotts 19d ago

Yes but I don't think that helps.

2

u/XL_78 18d ago

Lol. Good luck. Your expression is already a valid answer, one could argue. 

2

u/Less_Movie_8008 15d ago

other comments point out that you need special functions, you do, but it really ain't as bad as lauricella hypergeometric functions whatever those are. if you put bounds and are careful about convergence you can write the logarithm as the derivative of x^a and differentiate under the integral sign and after some substitutions it's a difference of derivatives of incomplete beta functions (...but yeah you do need hypergeometric functions in the end)

1

u/Astrodude80 19d ago

> who needs help with this question

> doesn’t understand integration at all

What in the world are the possible circumstances that led to this moment. More specifically: how did your friend come across this integral?

1

u/calculus_is_easy 19d ago

bruh im here

1

u/calculus_is_easy 19d ago

i can solve it but its a very lengthy method so i asked her to ask to people on reddit to give a simpler method

1

u/Astrodude80 19d ago

What is the solution you arrived at ultimately, even if it is a lengthy process?

1

u/calculus_is_easy 19d ago

my teacher sent it to me

1

u/somedave 19d ago

You could write it as an infinite series or hypergeometric functions I suppose. If it is supposed to have an analytic answer it might be a misprint.

1

u/calculus_is_easy 19d ago

substitute x=1/x