Recent content by Lodve

Hmmm not sure, what if a let the u equal to x^2 and derive it so that I get 2x?

Hi I'm struggling solving this integration: x^3 * sin x^2 My idea to solve this one is to apply integration by parts, but I can't get through using it. \frac{4x+2}{x^2(x^2+2)} this seems a hard nut to crack.

I struggle to solve this Differential equation: x^2y' +2xy = arctanx What I did was just divide x^2 on both side of the sign of equality in order to get the same form as a first-order linear diff.equation. After I've done that, I just multiply e^(2lnx) on the both side of the equation so...

Thank you for helping me :D Appreciate it ;)

Thank you for helping me :D

Sorry about that :P It won't happen again. I'm quite new here as you can see :P Yes I have derived both nominator and denominatorm and I got the same result as you got. However I still get "0/0", and it's much more difficult to derive both nominator and denominator again.

Bump!

Yes, I'm fimiliar with some the of standard table of integrals. I think I just change the equation above in the form \int \frac{1}{x^2+a^2} where a is an integer, so that the integration of the equation would likely be \frac{1}{a} arctan(x/a) But I'm not sure if I've done it right.

I know I have to apply l'hospital's rule to solve this equation, but it seems unsolvable in my eyes. I would be thankful if someone here could spend their time solving this one for me :D \frac{ln(cos2x)}{(tanx)^2} The variable x is closely 0.

I really struggle to integrate this fraction expression: \int \frac{2}{4x^2+1}

You know if you get inifinity over infinity you must "change" your fractionexpression by using L'hopitas rule. It says that if you either get infinity over ifinity or zero over zero you must derive nominator og denominator so that you get a fractionexpression which is capable of giving you an...