MHB Is It an $\arctan$, $\arccot$, or Something Else in This Integral Solution?

[Lorena_Santoro]

 

[skeeter]

partial fraction decomposition ... choice (b

 

[I like Serena]

Let's see if it is an $\arctan$ or an $\arccot$.
We can rewrite:
$$\frac{3}{x^2+x-2}=\frac{3}{(x+\frac 12)^2-\frac 14-2}=\frac{3}{(x+\frac 12)^2-\frac 94}$$
The minus sign in the denominator implies that it is neither $\arctan$ nor $\arccot$.

So the only possible answers are (b) and (d).
Let's check (b) by taking its derivative:
$$\frac d{dx}\left(\ln\left|\frac{x-1}{x+2}\right|+c\right) = \frac d{dx}(\ln|x-1|-\ln|x+2|)=\frac 1{x-1}-\frac 1{x+2}=\frac{(x+2)-(x-1)}{(x-1)(x+2)}=\frac{3}{x^2+x-2}$$
We have a match. Therefore (b) is the correct answer.

 

[Lorena_Santoro]

partial fraction decomposition ... choice (b

Yes, thanks!

 

[Lorena_Santoro]

Let's see if it is an $\arctan$ or an $\arccot$.
We can rewrite:
$$\frac{3}{x^2+x-2}=\frac{3}{(x+\frac 12)^2-\frac 14-2}=\frac{3}{(x+\frac 12)^2-\frac 94}$$
The minus sign in the denominator implies that it is neither $\arctan$ nor $\arccot$.

So the only possible answers are (b) and (d).
Let's check (b) by taking its derivative:
$$\frac d{dx}\left(\ln\left|\frac{x-1}{x+2}\right|+c\right) = \frac d{dx}(\ln|x-1|-\ln|x+2|)=\frac 1{x-1}-\frac 1{x+2}=\frac{(x+2)-(x-1)}{(x-1)(x+2)}=\frac{3}{x^2+x-2}$$
We have a match. Therefore (b) is the correct answer.

Thanks so much, especially for your time. You are perfectly right. Also, I liked how it started by doing 3=2+1 and then adding and negating x for factorization.