Integration By Partial Fractions

We have,



Equating the coefficient of and constant
terms, we get

Solving these equations, we get

Step 1: Expand the partial fraction form or use a strategic method.
We want Multiply throughout by : Step 2: Key approach to find .
Rather than solving for individually, note we only need the linear combination . A common trick: Evaluate the expression at certain convenient values or compare coefficients systematically. One direct trick is to rewrite: But a simpler approach might be to plug in a specific polynomial identity and pick off the relevant combination of coefficients. Alternatively, expanding and equating coefficients of e.g. might be straightforward. Step 3: Quick systematic expansion (outline).
Hence Step 4: Solve systematically or just find the required combination.
From , the second equation
Third:
Fourth:
Fifth: Step 5: Compute .
Substitute: Now let’s do partial fraction for that fraction only: Multiply both sides by : Now let’s expand quickly: Hence: So summing up: We match with . So:



Compute step by step:
Sum these partials: So
Now recall the entire expression has plus these fraction terms, so the final partial fraction is Hence in the original form: Finally: Now compute carefully: Yes, that matches the given correct answer of . Thus is the required value.

Step 1: Partial fraction decomposition.
We decompose the given rational function: By equating coefficients, solving for , and evaluating , we derive:

Step 1: Partial fraction decomposition.
Multiply both sides by :
Step 2: Find using .

Step 3: Expand and collect coefficients.

Step 4: Solve for , , .

Step 5: Compute .

We are given the partial fraction decomposition:
.
To solve this, let's make a temporary substitution . The expression becomes:
.
We can now decompose this simpler rational expression.
.
Using the cover-up method:
To find P, cover the term and substitute into the rest:
.
To find Q, cover the term and substitute into the rest:
.
So, the decomposition is .
Now, substitute back :
.
Comparing this with the given form :
For the first term, , which implies and .
For the second term, , which implies and .
The required value is .
.

Step 1: Understanding the Concept
This problem requires decomposing a rational function into its partial fractions. The denominator is already factored into a linear term and an irreducible quadratic term.
Step 2: Key Formula or Approach
To find the constants and , we first combine the fractions on the right side over a common denominator and then equate the numerators. We can find the constants by substituting convenient values of (like the root of the linear factor) and by comparing the coefficients of the powers of .
Step 3: Detailed Explanation
We start with the identity: Finding 'a' (Cover-up method): To find 'a', we can substitute the root of its denominator, which is . Finding 'b' and 'c' (Comparing coefficients): Expand the right side of the identity: Group the terms by powers of : Now, equate the coefficients of corresponding powers of on both sides. Coefficient of : Since , we get . Coefficient of : Since , we get . We can check our result with the constant term: Constant term: . . This matches, so our values are correct. Calculate the final expression: We need to find the value of . Step 4: Final Answer
The value of is 3.