0
$\begingroup$

This is the problem:

$$\int \frac{x+2}{x^2+x}$$

I am supposed to write $(y+2)$ as $\frac{1}{2}(2x+1)+\frac{3}{2}$

$$\frac{1}{2}\int \frac{2x+1}{x^2+x}+\frac{3}{2}\int \frac{3}{x^2+x}$$

Fine, is there supposed to be some way for me to know which fractions to use here or am I supposed to just guess or can I use any number here? What is this method?

This does not look like the integration by partial fraction method outlined in the book.

$\endgroup$
1
  • $\begingroup$ Hint: Expand $\frac{1}{2}(2x + 1) + \frac{3}{2}$ and see what happens. Then in the second integral, complete the square. $\endgroup$ Oct 1, 2017 at 6:07

2 Answers 2

4
$\begingroup$

Hint:

As $x^2+x=x(x+1),$

Set $x+2=Ax+B(x+1)=x(A+B)+B$

$\implies B=2$

$\endgroup$
3
$\begingroup$

You can do this : $$\frac {(x+1)+1}{x(x+1)}=\frac 1x + \frac 1{x(x+1)}=\frac 1x + \frac {(x+1-x)}{x(x+1)}=\frac 1x + \left(\frac 1x - \frac 1{x+1}\right)=\frac 2x - \frac {1}{x+1}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .