Author 
Message 
Prometheus


Original Member
Joined: Sun Aug 07, 2011 8:58 am Posts: 309

$ \displaystyle \int\frac{7x+3}{x^2+4x+5}dx$
Can't be split into partial fractions, and i can't solve it by recognition or substitution. Am i missing something or is there another way of integrating it (the above are the only i've been taught)?





DrRocket


Original Member
Joined: Fri Aug 05, 2011 2:22 am Posts: 477

Prometheus wrote: $ \displaystyle \int\frac{7x+3}{x^2+4x+5}dx$
Can't be split into partial fractions, and i can't solve it by recognition or substitution. Am i missing something or is there another way of integrating it (the above are the only i've been taught)? $ \displaystyle \int\frac{Mx+N}{Ax^2+2Bx+C}dx$ $ \displaystyle = \frac{M}{2C}lnA+2Bx+Cx^2 \ +$$ \displaystyle \frac {NCMB}{C\sqrt{ACb^2}} \ arctan(\frac{Cx+B}{\sqrt{ACb^2}}), \ \ [AC>B^2]$ $ \displaystyle = \frac {M}{2C}lnA+2Bx+Cx^2 +\frac{NCMB}{2C \sqrt{B^2AC}} \ ln$$ \displaystyle  \frac {Cx+B \sqrt{B^2AC}}{Cx+B + \sqrt {B^24AC}} , \ \ [AC<B^2]$ This seems a bit much for a homework problem in an introductory calculus class.
_________________ gone





Prometheus


Original Member
Joined: Sun Aug 07, 2011 8:58 am Posts: 309

DrRocket wrote: This seems a bit much for a homework problem in an introductory calculus class. I'm glad i'm not the only one who thinks so. Maybe they just want us to recognise the limitations of what we have been taught so for. I'll work through the general solution and try to find why it works. Thanks.





Prometheus


Original Member
Joined: Sun Aug 07, 2011 8:58 am Posts: 309

I made it $ \displaystyle \frac{7}{2}lnx^2+4x+511acrtanx+C$
This is about the hardest we have to deal with on our syllabus. I actually found this one easier than integrating with trig. substitutions (assuming i got this one right). Thanks for the help.





x(xy)


Original Member
Joined: Sat Aug 06, 2011 3:44 pm Posts: 298 Location: UK

I just tried integration by parts to try to solve this but ended up getting horribly complicated terms which I have no wish to integrate! Anyway, is there any way that the fraction could be reduced down by a denominator heavy polynomial division I know how to do it if the top has higher index powers than the bottom e.g. if the top is a quartic equation and the denominator is a quadratic then this would reduce to a quadratic equation plus a linear one over the original denominator but I'm not sure if the same can be done with this type of fraction?
Anyway, I'll think about this problem I hate not being able to do a question.
P.S. I know that DrRocket's method is probably correct, but I haven't got to that level of calculus yet so I don't really understand it at the moment, I'm just wondering if it can be solved through one of the ways I (or you) know i.e. by parts, substitution, partial fractions, polynomial division etc...
_________________ "Nature doesn't care what we call it, she just does it anyway".  Feynman





Prometheus


Original Member
Joined: Sun Aug 07, 2011 8:58 am Posts: 309

The question was in a set a very similar questions, but all requiring different methods for solving. I suspect, then that this cannot be solved in the standard ways we know. This is at the limit of my understanding, but this is how i've gone about it  it should work for any linear over quadratic (of $ \displaystyle D<0$) function, assuming you can make a matching differential.
1st i took $ \displaystyle \frac{7}{2}$ outside the integral leaving $ \displaystyle 2x+4$, a matching differential of $ \displaystyle x^2+4x+5$. Once this is done it's quite easy to get $ \displaystyle \frac{7}{2}ln(x^2+4x+5)$.
But then there's also a 14 left from $ \displaystyle \frac{7}{2} \times 4$. To get it back to the orginal 3, we need 11, giving: $ \displaystyle \int\frac{11}{x^2+4x+5}dx$
Then, when the quadratic is in completed square form $ \displaystyle (x+2)^2+1$, we can see we can make the substitution $ \displaystyle (x+2)^2 = tan^2\theta$. From that, i eventually got (maybe incorrectly, now i look at Dr. Rockets formula) $ \displaystyle 11arctanx$
Hope this doesn't count as breaking the rules, since i'm solving my own homework.





Prometheus


Original Member
Joined: Sun Aug 07, 2011 8:58 am Posts: 309

It was the right answer, though there was an element of luck involved in getting there.





