Question

Evaluate the following integrals: $$ \int \frac{x^{2}+2 x-1}{2 x^{2}+3 x+1} d x $$

Short answer

Question: Evaluate the integral of the function \(f(x) = \frac{x^2 + 2x - 1}{2x^2 + 3x + 1}\). Answer: The integral of the function \(f(x)\) is given by \(-\frac{3}{4} \ln{|2x + 1|} + C\), where C represents the integration constant.

Step-by-step solution

Find the factors of the denominator

Factor the polynomial in the denominator, \(2x^2 + 3x + 1\). We look for two binomials of the form \((ax + b)(cx + d)\), which when multiplied together, give us the original polynomial. Since the coefficient of \(x^2\) is 2, we know that either \(a=c=1\) or \(a=c=2\) and \(b=d=1\). Trying the latter option, we find \((2x + 1)(x + 1) = 2x^2 + 2x + 1x + 1 = 2x^2 + 3x + 1\). Thus, the denominator can be factored as \((2x + 1)(x + 1)\).

Decompose into partial fractions

We want to find two constants, A and B, such that $$ \frac{x^2 + 2x - 1}{(2x + 1)(x + 1)} = \frac{A}{2x + 1} + \frac{B}{x + 1}. $$ To do this, we clear the denominators by multiplying both sides by \((2x + 1)(x + 1)\). This gives us: $$ x^2 + 2x - 1 = A(x + 1) + B(2x + 1). $$ If we solve for A and B, we will find the decomposition of our integrand.

Solve for A and B

We want to satisfy the equation \(x^2 + 2x - 1 = A(x + 1) + B(2x + 1)\) for every value of x. One way to do this is to plug in values for x that will eliminate either A or B. When \(x = -1\), we get: $$ (-1)^2 + 2(-1) - 1 = A(-1 + 1) + B(2(-1) + 1) \Rightarrow 0 = -B, $$ which gives us \(B = 0\). When \(x = -\frac{1}{2}\), we get: $$ \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 1 = A\left(-\frac{1}{2} + 1\right) + 0\left(2\left(-\frac{1}{2}\right) + 1\right) \Rightarrow -\frac{3}{4} = \frac{1}{2}A, $$ which gives us \(A = -\frac{3}{2}\). Now we have the decomposition: $$ \frac{x^2 + 2x - 1}{(2x + 1)(x + 1)} = \frac{-\frac{3}{2}}{2x + 1} + \frac{0}{x + 1} = -\frac{3}{2(2x+1)}. $$

Integrate the decomposed function

Now we can integrate the decomposed function: $$ \int \frac{x^2 + 2x - 1}{(2x + 1)(x + 1)} d x = \int -\frac{3}{2(2x+1)} dx. $$ Let \(u = 2x + 1\), then \(du = 2 dx\), and we find the integral: $$ -\frac{3}{4} \int \frac{1}{u} du = -\frac{3}{4} \ln{|u|} + C = -\frac{3}{4} \ln{|2x + 1|} + C. $$ So the integral evaluates to $$ -\frac{3}{4} \ln{|2x + 1|} + C. $$