Science:Math Exam Resources/Courses/MATH101/April 2013/Question 06 (b)
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Question 06 (b) 

FullSolution Problem. Justify your answers and show all your work. Simplification of numerical answers is required. Evaluate 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Try completing the square in the denominator, then use a substitution. 
Hint 2 

(alternate) Is there a simple substitution? If not, can you modify the integrand so that there is? 
Hint 3 

What is the derivative of the denominator? How does it compare to the numerator? 
Hint 4 

Don't forget the constant! 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 1 

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Please rate my easiness! It's quick and helps everyone guide their studies. Let's complete the square in the denominator. This suggests using the substitution , so that and . Then For the first integral, we use the substitution , so that . For the second integal, this looks like an integral, but we need to scale. We use the substitution , so that and . Then Altogether, with the arbitrary constant of integration C_{3}. 
Solution 2 

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Please rate my easiness! It's quick and helps everyone guide their studies. Note that this is already in reduced form using partial fraction decomposition. We see that the numerator is almost the derivative of the denominator. We can modify it to be exactly the derivative by subtracting 1, then adding 1 to make up for it. Then using a simple substitution in the first integral and completing the square in the second, 