Recent content by DragonIce

well i did that... if you look carefully you can notice that i treat t=arctan(x) so basicaly \int e^t\frac{t}{1+x^2}dx equal to \int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx and my answer is v=\arctan(x)*e^{\arctan(x)}-\frac{e^{\arctan(x)}}{x^2+1}+C and then i turn back to y=u*v and...

Anyway if you are right i still don't know what to do with arctan pi/4-1+C/e^arctan pi/4=0 C=e^arctan x - arctan x * e^arctan x I have to transform it somehow to get rational answer

(u*v)'=u'*v+u*v' I meant it like dt=t' I should write ' instead of d, but i am right here i think. If u=f(x) and v=e^q(x) Then (u*v)'=f'(x)*e^q(x)+f(x)*q'(x)*e^q(x) It doesn't solve my problem with Constant

Homework Statement \frac{dy}{dx}+\frac{y}{(1+x^2)} = \frac{\arctan x}{(1+x^2)} when y(\frac{\pi}{4})=0 Homework Equations \frac{dy}{dx}+Q(x)*y=F(x) The Attempt at a Solution y=u*v u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)} \frac{du}{dx}+\frac{u}{(1+x^2)}=0...