Q.Integrate the following function:
We decompose the rational function into simpler partial fractions, integrate each term using the natural logarithm rule, and combine the result. The integral is .
The key idea here is Partial Fraction Decomposition. When you have a rational function (a polynomial divided by another polynomial) and the denominator factors into distinct linear factors, you can break the complicated fraction into a sum of simpler fractions — each with a single linear denominator. This turns a messy integration problem into a sum of easy logarithmic integrals.
Why does this work? The denominator is already factored. The numerator is of lower degree than the denominator, so we can directly write:
where and are constants we need to find. Once we find them, integrating gives , and similarly for the other term.
Let’s find and step by step.
- Set up the equation. Multiply both sides by the denominator to clear fractions:
This identity must hold for all .
- Solve for and . There are two efficient methods. I’ll use the substitution method (also called the cover-up method) because it’s fastest for distinct linear factors.
- To find , set (this makes the term vanish because ):
- To find , set (this makes the term vanish):
The cover-up method works because plugging in the root of a factor isolates the corresponding constant. For , cover and substitute into the rest: , so . Similarly, cover and substitute : , so . This is a huge time-saver in exams.
- Write the decomposed form. Substituting and back:
- Integrate term by term. Now the integral becomes:
Each term is of the form . Here for both, so:
A common mistake is forgetting the absolute value signs inside the logarithm. Since the domain of the original function excludes and , the integral is defined on intervals not containing these points, so absolute values are necessary for the general antiderivative.
- Simplify if desired. Using logarithm properties, , so we could also write:
But the form is perfectly acceptable and often preferred.
The integral is .
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