Question

Use the approaches discussed in this section to evaluate the following integrals. $$\int \frac{x-2}{x^{2}+6 x+13} d x$$

Short answer

Question: Evaluate the following integral: $$\int \frac{x-2}{x^2+6x+13} dx$$ Answer: $$\int \frac{x-2}{x^2+6x+13} dx = -\frac{5}{4} \ln|x+3| + 2 \arctan\left(\frac{x+3}{2}\right) + C$$

Step-by-step solution

Rewrite the integrand as a partial fraction decomposition

Write the integrand as a sum of two simpler fractions: $$\frac{x-2}{x^2+6x+13} = \frac{A}{x+3} + \frac{B}{(x+3)^2+4}$$

Find A and B

Clear the denominators by multiplying both sides of the equation by \((x^2+6x+13)\): $$(x-2) = A[(x+3)^2+4] + B(x+3)$$ Now, we can plug in some values of x to find A and B. By setting \(x=-3\), we get: $$-5 = A(4)$$ Solve for A: $$A = -\frac{5}{4}$$ Now, differentiate the right side with respect to x to find B: $$1 = 2A(x+3) + B$$ Insert the value of A we found earlier: $$1 = 2\left(-\frac{5}{4}\right)(x+3) + B$$ Now, again set \(x = -3\): $$1 = B$$ Now, we have \(A=-\frac{5}{4}\) and \(B=1\). So, we can rewrite the integrand as: $$\int \frac{-\frac{5}{4}}{x+3} + \frac{1}{(x+3)^2+4} dx$$

Perform a substitution

Substitute \({u} = x+3\) and \(du = dx\) to simplify the problem. The integral becomes: $$\int \frac{-\frac{5}{4}}{u} + \frac{1}{u^2+4} du$$

Split the integral

Separate the integral into two parts: $$-\frac{5}{4}\int \frac{1}{u} du + \int \frac{1}{u^2+4} du$$

Integrate each part

Integrate the first part: $$-\frac{5}{4} \ln|u| + C_1 = -\frac{5}{4} \ln|x+3| + C_1$$ For the second part, rewrite \(\int \frac{1}{u^2+4} du\) as \(\int \frac{1}{(\frac{u^2}{4})+(1)} du\). Now, substitute \({v} = \frac{u}{2}\), so \(dv = \frac{1}{2} du\). The integral becomes: $$\int \frac{1}{v^2+1} (2 dv) = 2 \int \frac{1}{v^2+1} dv = 2 \arctan(v) + C_2 = 2 \arctan\left(\frac{x+3}{2}\right) + C_2$$

Combine the integrals and simplify the result

Add the two integrals and their constants: $$-\frac{5}{4} \ln|x+3| + 2 \arctan\left(\frac{x+3}{2}\right) + C$$ So, the final solution is: $$\int \frac{x-2}{x^2+6x+13} dx = -\frac{5}{4} \ln|x+3| + 2 \arctan\left(\frac{x+3}{2}\right) + C$$