Compute $\int \frac{-3}{2x^2-x-1} dx$

Question

How to compute $\int \frac{-3}{2x^2-x-1}\ dx$?

I tried completing the square

and ended with :

$\frac{-3}{(2(x-1/4)^2-9/8)}$

Then I did U-sub for $u = x-1/4$

and got

$-3 \cdot 8 \cdot \int 1/(16u^2)-9 du$

What do I do from here?

Answer 1

Now you do $u=\frac34t$ and $\mathrm du=\frac34\mathrm dt$. Then your integral becomes$$-\frac9{32}\int\frac1{9t^2-9}\,\mathrm dt=-\frac1{32}\int\frac1{t^2-1}\,\mathrm dt.$$And now you write$$\frac1{t^2-1}=\frac12\left(\frac1{t-1}-\frac1{t+1}\right).$$

Answer 2

You can solve this with partial fractions.

$\frac 6 {2x-1} + \frac {-3}{x-1} = \frac {-3} {2x^2-x-1}$

Then you can integrate each fraction separately; they'll both result in logs.

It might be instructive to compare the result to what you get with José Carlos Santos' answer to see whether they're the same. BTW, you didn't include the limits in your work, but when do substitution, you have to change the limits accordingly, so you should keep track of them.

Answer 3

$$\frac{-3}{2x^2-x-1}=\frac{-3}{(2x+1)(x-1)}=\frac{2}{2x+1}-\frac{1}{x-1},$$ which gives $$\ln\left|x+\frac{1}{2}\right|-\ln|x-1|+C.$$

Answer 4

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

$$=\int (\frac{2}{2x+1} - \frac{1}{x-1})$$

$$= \int \frac{2}{2x+1} - \int \frac{1}{x-1}$$

$$= \ln |2x+1| - \ln|x-1| + C$$