How do you integrate #(x^2 + 2x -5) / [(x - 2)*(x+1)*(x^2+1)]#?

Question

Emilia Aguilar · Accepted Answer

#1/5 ln |x - 2| + ln |x+1| + 8/5 arc tan x - 3/5 ln |x^2 + 1| + C#

I am not 100% sure for this one but I have attached a picture so that you may see what I did.
Please, feel free to ask any questions and even more, to correct what I did wrong :)
Integrals can be a pain lol

Emilia Aguilar · Answer

undefined To integrate $(x^2 + 2x -5) / [(x - 2)(x+1)(x^2+1)]$, you first need to perform partial fraction decomposition to express the rational function as a sum of simpler fractions. Then, you can integrate each term separately. The partial fraction decomposition of the given expression would be:

$$\frac{x^2 + 2x -5}{(x - 2)(x+1)(x^2+1)} = \frac{A}{x - 2} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}$$

Solving for $A$, $B$, $C$, and $D$ by equating coefficients, you'll find:

$A = \frac{3}{10}$, $B = -\frac{3}{10}$, $C = \frac{1}{2}$, $D = -\frac{1}{2}$

So, the integral becomes:

$$\int \frac{3}{10(x - 2)} - \frac{3}{10(x + 1)} + \frac{\frac{1}{2}x - \frac{1}{2}}{x^2 + 1} \,dx$$

Now you can integrate each term separately:

$$\frac{3}{10}\ln|x - 2| - \frac{3}{10}\ln|x + 1| + \frac{1}{4}\ln(x^2 + 1) + C$$

where $C$ is the constant of integration.