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Question 7
3. (a) Express \( \frac{5}{(x-1)(3x+2)} \) in partial fractions. (b) Hence find \( \int \frac{5}{(x-1)(3x+2)} \, dx \), where \( x > 1 \). (c) Find the particu... show full transcript
Step 1
Answer
To express ( \frac{5}{(x-1)(3x+2)} ) in partial fractions, we set up the equation:
[ \frac{5}{(x-1)(3x+2)} = \frac{A}{x-1} + \frac{B}{3x+2} ]
Multiplying through by ( (x-1)(3x+2) ) gives:
[ 5 = A(3x+2) + B(x-1) ]
Substituting convenient values of ( x ):
Thus, the partial fraction decomposition is:
[ \frac{5}{(x-1)(3x+2)} = \frac{1}{x-1} - \frac{3}{3x+2} ]
Step 2
Answer
Using the partial fraction result, we can integrate:
[ \int \frac{5}{(x-1)(3x+2)} , dx = \int \left( \frac{1}{x-1} - \frac{3}{3x+2} \right) , dx ]
This results in:
[ = \ln |x-1| - \ln |3x+2| + C = \ln \left( \frac{x-1}{3x+2} \right) + C ]
Step 3
Answer
We first rewrite the equation:
[ \frac{dy}{y} = \frac{5}{(x-1)(3x+2)} , dx ]
Integrating both sides:
[ \ln |y| = \int \frac{5}{(x-1)(3x+2)} , dx ]
Substituting our previous solution:
[ = \ln \left( \frac{x-1}{3x+2} \right) + C ]
Exponentiating both sides gives:
[ y = K \cdot \frac{x-1}{3x+2} ]
Using the condition ( y = 8 ) at ( x = 2 ):
[ 8 = K \cdot \frac{2-1}{3(2)+2} \Rightarrow 8 = K \cdot \frac{1}{8} \Rightarrow K = 64 ]
Thus, the particular solution is:
[ y = \frac{64(x-1)}{3x+2} ]
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