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The integral evaluates to
Solución
Write the integrand as
Since the numerator is a degree 4 polynomial and the denominator is degree 3, we perform long division. Dividing
Multiply the divisor by
Subtract from the numerator:
Next, divide the new leading term
Multiply the divisor by
Subtract:
The division yields:
Thus, the integral becomes:
Focus on the partial fraction for
Assume
Multiply both sides by
Expand the right-hand side:
Equate coefficients:
- Coefficient of
: . - Coefficient of
: . - Constant term:
.
Thus, we have:
The decomposition is:
The integral now is:
- Evaluate
:
So,
- Evaluate
:
- Evaluate
in two parts:
(a) For
Let
(b) For
Summing all parts, the integral is:
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Simplifique esta solución Mind Expander
To solve the integral
we can begin by breaking down the fraction into partial fractions. We express the integrand as:
where
Simplifying this gives us:
Now, we can match coefficients with the polynomial on the right:
- For
: There is no term, so its coefficient must be , - For
: The coefficient must be , - For
: The coefficient must be , - For
: The coefficient must be , - For the constant term: It must be
.
From this, we can set up a system of equations:
- Coefficient of
: (no restrictions) - Coefficient of
: (no restrictions) - Coefficient of
: , - Coefficient of
: , - Constant term:
.
From
Now we have
Next, we can integrate term by term:
The first integral is straightforward:
For the second integral, we can break it down further:
The integral of the first part can be solved using substitution
The second integral is simply:
Putting it all together, we have:
where
Thus, the final answer is:
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