Grueling problem. Satisfying solution.

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[@hiyayahiyaya5645]

bro, just break ln[(a²+x²)/(b²+x²)] into ln(a²+x²)-ln(b²+x²) before finding its derivative, here is no need to differentiate it and do partial fractions .😂😅

[@alvargd6771]

-studies 69 billion integration tecniques -doesnt know basic log rules

[@GangulaArjunNarenReddy]

awsome vid but there are quicker and tidier ways

[@phill3986]

Since it was a "pure" quadratic, you could have subbed a variable for x² during the partial fraction evaluation. That would have made it linear, making the partial fractions easier

[@onegreengoat9779]

Very nice. It sounds like I did a similar approach to others in the comment section. I first looked at the antiderivative of ln(a^2+x^2). I noticed the pattern of "a^2+x^2" to make a trigonometric (tangent) substitution. This gave me an antiderivative of x*ln(a^2+x^2)-2x+2*|a|*arctan(x/|a|). Now you can subtract off an expression with b replacing a to get the antiderivative of the original function to be x*ln((a^2+x^2)/(b^2+x^2)) + 2*|a|*arctan(x/|a|) - 2*|b|*arctan(x/|b|). To solve the improper integral, you can take a definite integral from 0 to N, then take the limit as N gets large. You would use L'Hopitâl's rule and the properties of arctan to derive the final answer, just like yours. pi*(|a|-|b|). Very fun problem.

[@owl3math]

Hi Kyle. Nice approach! And thanks! 😁

[@chriscobe1990]

Very nice solution. I take the liberty to suggest a modification to the conditions of the constants a and b; in particular, they only have to b greater than 0, not equal to, since at the end of the calculation we find 1/a and 1/b

[@alexkaralekas4060]

Or just break the ln() into two ln(a²+x²)-ln(b²+x²) and just calculate one integral with di method but it would be much easier and you need to just calculate one of the ln() cause the other one is the same

[@adandap]

Yes, that's what I did, but you need a little work with limits when you do it that way. I integrated over [0,L] and then used arctan[L/a] = pi/2 - arctan[a/L] ~ pi/2 - a/L for large L to regularise the infinite integrals.

[@owl3math]

Yep there are some shortcuts with tradeoffs you can do. I think there are some quicker ways.

[@nuranichandra2177]

Nice and intimidating integrals

[@owl3math]

Yep agree on that! thanks

[@txikitofandango]

I wonder, since it's ln(something + f(x)), would a Maclaurin series work here

[@owl3math]

Yes I think it would work

[@banjo2402]

Really nice, how do you know how to set up the partial fraction decomposition with Ax+B and Cx+D and not just A and B?

[@ben_adel3437]

you do one less degree of the denominator so if it's a x³+... you'll write Ax²+Bx+C etc

[@owl3math]

Thanks! And see comment from Ben below to setup numerator with one less degree than the denominator :)

[@banjo2402]

@@ben_adel3437 thanks

[@felipefred1279]

Nice problem

[@owl3math]

Thanks Felipe! :) have a good day

[@user-lu6yg3vk9z]

@@owl3mathvideo requests integral of 1/(1+sin(x)+cos(x)) dx Hint: Use this method to substitution Cos(pi/4-x)

[@Babyshark-co8ks]

Feynman trick is so much easier

[@Samir-zb3xk]

True, i solved it very quickly using feynman integration

[@owl3math]

I think that’s what UK integration bee recommends for this one