Can someone please help me with this question. I have no idea where to start and also if I got started I wouldn't know what to do.
Evaluate the integral
∫(1x−3)/(x^2+2x+4)^2 dx
Note: Use an upper-case "C" for the constant of integration.
This one is pretty tricky. Take a look at this handy integral calculator, which shows all the details of the manipulations involved.
https://www.integral-calculator.com
Just type in your function, using parentheses for clarity.
To evaluate the given integral, you can use a technique called partial fraction decomposition. This involves breaking down the integrand into simpler fractions which will make integration easier.
To begin, let's factor the denominator of the fraction:
x^2 + 2x + 4 = (x + 1)^2 + 3
Now, we can rewrite the integrand using partial fraction decomposition:
(1x - 3)/(x^2 + 2x + 4)^2 = A/(x^2 + 2x + 4) + B/(x^2 + 2x + 4)^2
To find the values of A and B, we need to find a common denominator and equate the numerators:
1x - 3 = A(x^2 + 2x + 4) + B
Expanding the equation and collecting like terms, we get:
1x - 3 = Ax^2 + 2Ax + 4A + B
Now, equating the coefficients of like terms on both sides, we have the following system of equations:
1 = A
-3 = 2A + 4A + B
From the first equation, we find A = 1. Substituting A = 1 in the second equation, we get:
-3 = 2 + 4 + B
-3 = 6 + B
B = -9
Now that we have A and B, we can rewrite the given integral as:
∫(1x−3)/(x^2+2x+4)^2 dx = ∫(1/(x^2 + 2x + 4) + (-9)/(x^2 + 2x + 4)^2) dx
To evaluate this integral, we can now use the power rule of integration and the fact that the integral of the form ∫1/(x^2 + a^2) dx = (1/a) * arctan(x/a) + C. Additionally, the integral of -(2/a^3)/(x^2 + a^2) dx = -(1/a^2) * 1/(x^2 + a^2) + C, where C is the constant of integration.
Applying these rules, the integral becomes:
∫(1/(x^2 + 2x + 4) + (-9)/(x^2 + 2x + 4)^2) dx = arctan(x + 1)/√3 - 3/(x^2 + 2x + 4) + C
So, the evaluated integral is:
∫(1x−3)/(x^2+2x+4)^2 dx = arctan(x + 1)/√3 - 3/(x^2 + 2x + 4) + C