The value of \(\int \frac {x^2}{x^2 - 3x + 2}dx\) will be ___________, where c is an arbitrary constant.
x - log |x - 1| + 4 log |x - 2| + c
Integral Calculation: Solving Rational Functions
This solution details the step-by-step process to evaluate the indefinite integral of the given rational function, \(\int \frac{x^2}{x^2 - 3x + 2}dx\). We will employ techniques such as polynomial long division and partial fraction decomposition to simplify the integrand before integration.
Understanding the Rational Function Integral
The given integral is of a rational function where the degree of the numerator (\(x^2\), degree 2) is equal to the degree of the denominator (\(x^2 - 3x + 2\), degree 2). When the degree of the numerator is greater than or equal to the degree of the denominator, the first step is always to perform polynomial long division or manipulate the numerator to simplify the expression.
The denominator can be factored as follows:
\(x^2 - 3x + 2 = (x-1)(x-2)\)
Now, let's rewrite the numerator \(x^2\) in terms of the denominator:
We can add and subtract terms to make the numerator resemble the denominator:
\[ \frac{x^2}{x^2 - 3x + 2} = \frac{x^2 - 3x + 2 + 3x - 2}{x^2 - 3x + 2} \]
This allows us to split the fraction:
\[ = \frac{x^2 - 3x + 2}{x^2 - 3x + 2} + \frac{3x - 2}{x^2 - 3x + 2} \]
\[ = 1 + \frac{3x - 2}{(x-1)(x-2)} \]
So, the integral becomes:
\[ \int \frac{x^2}{x^2 - 3x + 2}dx = \int \left(1 + \frac{3x - 2}{(x-1)(x-2)}\right) dx \]
\[ = \int 1 dx + \int \frac{3x - 2}{(x-1)(x-2)} dx \]
The integral of \(1\) is simply \(x\).
Partial Fraction Decomposition
Next, we need to evaluate the integral of the remaining rational term: \(\int \frac{3x - 2}{(x-1)(x-2)} dx\). Since the denominator consists of distinct linear factors, we can use partial fraction decomposition.
Let:
\[ \frac{3x - 2}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \]
To find the constants \(A\) and \(B\), we multiply both sides by \((x-1)(x-2)\):
\[ 3x - 2 = A(x-2) + B(x-1) \]
Calculating Constants A and B
- To find A: Set \(x = 1\) (the root of \(x-1\)):
\[ 3(1) - 2 = A(1-2) + B(1-1) \]
\[ 1 = A(-1) + B(0) \]
\[ 1 = -A \implies A = -1 \]
- To find B: Set \(x = 2\) (the root of \(x-2\)):
\[ 3(2) - 2 = A(2-2) + B(2-1) \]
\[ 6 - 2 = A(0) + B(1) \]
\[ 4 = B \]
So, the partial fraction decomposition is:
\[ \frac{3x - 2}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{4}{x-2} \]
Final Integration Steps
Now, we substitute this back into our integral:
\[ \int \frac{3x - 2}{(x-1)(x-2)} dx = \int \left(\frac{-1}{x-1} + \frac{4}{x-2}\right) dx \]
\[ = \int \frac{-1}{x-1} dx + \int \frac{4}{x-2} dx \]
Using the standard integral \(\int \frac{1}{u} du = \log|u| + C\):
\[ = -\log|x-1| + 4\log|x-2| + C_1 \]
Combining all parts of the original integral:
\[ \int \frac{x^2}{x^2 - 3x + 2}dx = \int 1 dx + \int \frac{3x - 2}{(x-1)(x-2)} dx \]
\[ = x + (-\log|x-1| + 4\log|x-2|) + C \]
\[ = x - \log|x-1| + 4\log|x-2| + C \]
where \(C\) is the arbitrary constant of integration.
Comparing with Options
Let's compare our derived solution with the provided options:
| Option | Expression |
|---|---|
| 1 | \(x^2 + e^{2x} + 2 \log x + c\) |
| 2 | \(x - \log |x - 1| + 4 \log |x - 2| + c\) |
| 3 | \(x + e^x + \log |x + 1| + c\) |
| 4 | \(1 + x^2 \log |x + 1| - e^x + c\) |
Our result, \(x - \log|x-1| + 4\log|x-2| + C\), perfectly matches Option 2.
Important Questions from Indefinite Integrals
What is ∫ (e log x + sin x) cos x dx equal to?
Evaluation of \(\displaystyle\int{\dfrac{1-\tan x}{1+\tan x}}\ dx\) is:
The value of \(\int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} \) is:
The value of \(\int_0^\pi {{{\sin }^6}} x \cdot {\cos ^5}x dx\) is:
- \(\rm \int\frac{2x+3}{x^3+x^2-2x}dx=?\)
