Example 103 Integrate 1 5 x 2 4 Solution Now we see the ax 2 b form so we will

Example 103 integrate 1 5 x 2 4 solution now we see

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Example 10.3. Integrate 1 5 x 2 - 4 . Solution. Now we see the ax 2 - b form, so we will let x be something times sec( θ ). To make it easier to work with, factor out a 5 from the denominator: Z dx q 5 ( x 2 - 4 5 ) = 1 5 Z dx q x 2 - 4 5 . Now, we will let x = 2 5 sec( θ ), so that x 2 - 4 5 = 4 5 sec 2 ( θ ) - 4 5 = 4 5 (sec 2 ( θ ) - 1) = 4 5 tan 2 ( θ ) . Don’t forget dx = 2 5 sec( θ ) tan( θ ) . Thus, the integral becomes 1 5 Z 2 5 sec( θ ) tan( θ ) 2 5 tan( θ ) = 1 5 Z sec( θ ) = 1 5 ln | sec( θ ) + tan( θ ) | + C. Finally, we want this back in terms of x . Write sec( θ ) = 5 x 2 , so the triangle looks like: 2 5 x 2 - 4 5 x θ Here we used that sec( θ ) is hypotenuse/adjacent, and got the remaining side by Pythagorean theorem. So sec( θ ) = 5 2 x (which we knew already), and tan( θ ) = 5 x 2 - 4 2 . Thus, our answer is 1 5 ln 5 2 x + 1 2 p 5 x 2 - 4 + C . 20
Remark. The trickiest part of trig sub is knowing what to make x , so just make sure you understand what identities you are using for each of the three types demonstrated above. 11 Partial Fractions (8.5) We begin with words of caution: CAUTION. The first, and most important, thing to note about partial fractions is that you only use it when you can factor the denominator. For example, the integral Z dx x 2 - 2 x + 6 is not a partial fractions problem because the denominator cannot be factored (such things are called irre- ducible ). CAUTION. The second important point is that even if the denominator is factorable, you need the degree of the numerator to be strictly less than the degree of the denominator. If it is not, you must do long division first. Now that we have made note of these two facts, we can actually describe the method. The idea is that we would like to break down fractions such as 1 x 2 - x into easier pieces: 1 x ( x - 1) = - 1 x + 1 x - 1 . We will do this by making a guess, i.e. assuming we can write 1 x ( x - 1) = A x + B x - 1 , and then solving for A and B . However, first you need to learn what to guess. There are several cases, all having to do with how the denominator factors. For example, for an integral like Z dx ( x - 1) 2 ( x 2 + 1) 2 , the key numbers you want to pay attention to are: the degree of each irreducible term (in this case, the x - 1 and the x 2 + 1), and the powers they are raised two (in this case, both 2’s). You will then decompose them factor by factor. Let’s look at the cases: (1) Linear factor to the first power : If you had something like x - 1 in the denominator, then all you need in the decomposition is constant x - 1 . So, for example, we wanted to break down 1 ( x - 2)( x - 3) , our guess would be A x - 2 + B x - 3 , since both x - 2 and x - 3 are linear, and each is raised only to the first power. (2) Linear factors to a higher power : If you have something like ( x + 1) 2 in the denominator, then the fact that x + 1 is still only linear tells us we only need constants in the numerator. However, we must account for all powers of ( x + 1), so we would need to have A x +1 + B ( x +1) 2 . An example: if we had 1 ( x - 2) 2 ( x - 3) , we would notice we still have linear terms, but now the x - 2 is squared. We deal with this by just remembering to account for every power of our linear terms. So we would have to break

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