Integral Help: Solve \frac {2(1+x)} {1-2x-x^2} dx

[Sparky_]

Greetings

Can you help with the following integral:

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

I'm reasonably sure my setup is correct up to this integral. I tried to factor and do some canceling. - no luck'

thoughts and direction

Thanks
-Sparky-

 

[neutrino]

Have you tried substitution?

 

[ssd]

Put, x^2+2x-1=z

 

[Sparky_]

AHHH! thanks -

$$u = (1-2x-x^2)$$
$$du = -2 - 2x dx$$
$$dx = \frac {du} {-2(1+x)}$$
$$-\int \frac {-du} {u}$$

$$= ln(1-2x-x^2)$$

This solution is in the exponent of "e"

and leads to the integral below.
Question: can you suggest a start for:

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

I've tried various substitutions again and don't see it.

I've tried $$u = -x^2 - 2x$$
$$du = -2x - 2 dx$$
or
$$(-2(x+1) )dx$$
$$dx = \frac {du} {-2(x+1)}$$

leaves me with a (x+1) term

thanks
Sparky_

 

[pki15]

Note that $$1-2x-x^2 = -(x+1)^2 + 2$$.

Now separate, and integrate.

This is completing the square. Also you could multiply the bottom out and long divide to get a similar result.