Francis Woodhouse	Posted on Saturday, 21 June, 2003 - 02:36 pm: I'm having problems getting the correct answer for Question 3 in exercise 13.1 of "Teach Yourself Calculus". The question is to find The book gives the answer as: My answer, however, is: I got my answer by first performing polynomial long division, which gave two fractions which were easy to integrate. However, it appears that either I keep going wrong somewhere, or the book is wrong. Can anyone get the answer the book has, and if so, how? Thanks!
Alice Thompson	Posted on Saturday, 21 June, 2003 - 02:48 pm: I agree with them, but don't get the a/b2 term, but it can be part of the constant. You can rewrite the fraction as: . When you integrate , you must divide by b, giving their answer.
Kerwin Hui	Posted on Saturday, 21 June, 2003 - 02:59 pm: To get the a/b2 term: Let u=a+bx, then we have du/dx=b, x=(u-a)/b, so Kerwin
Francis Woodhouse	Posted on Saturday, 21 June, 2003 - 03:13 pm: Aha, substitution. Performing the substitution myself, I got the same answer as the book. It's a bit odd that substitution is required to get the exact same answer in a chapter all about integration of algebraic fractions... haven't had to use a substitution in all of the other questions in the exercise.
Mark Durkee	Posted on Saturday, 21 June, 2003 - 03:24 pm: You can do it without substitution using Alice's method above: which can be integrated.
Francis Woodhouse	Posted on Saturday, 21 June, 2003 - 04:07 pm: Turns out that my initial answer was different from the answer I was getting the rest of the time - should've been 1 / b2 , not 1 / b, outside the brackets. And yes, you can get the answer with Alice's method - however, as she says, it's missing the a/b2 term. Although this could be said to be in the constant of integration, a second integration would be wrong if it was missing the a/b2 term. I think, anyway. Thanks, you three - the help is appreciated.
Philip Ellison	Posted on Saturday, 21 June, 2003 - 04:18 pm: The second integration would not be affected, as ò(a/b2 + C1)dx=(a/b2 + C1)x and òC2 dx=C2 x, i.e. in both cases a second integration results in an arbitrary constant multiplied by x.
Francis Woodhouse	Posted on Saturday, 21 June, 2003 - 04:25 pm: Agh, silly me. Forgot about integrating the first constant.
Chris Tynan	Posted on Saturday, 21 June, 2003 - 06:54 pm: I think we also have to consider the case when b=0? The integral is then trivial, as long as a is not 0, but i think it should still be dealt with.