$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $ substitution

Question

I have this integral. $$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $$ But I am unable to do it right and I just don't know where is the problem in my steps.

My steps:

Complete the square $$\int \frac{1}{{2-(x+1)^2}} \, \mathrm{d}x $$ $$\frac{1}{2}*\int \frac{1}{{-(x+1)^2}} \, \mathrm{d}x $$ Substitute $$t=x+1$$ $$\frac{1}{2}*\int \frac{1}{{-t^2}} \, \mathrm{d}x $$ $$\frac{1}{2}*\int {{-t^{-2}}} \, \mathrm{d}x $$ $$\frac{1}{2}* \frac{-t^{{-1}}}{{-1}}+c $$ $$ \frac{1}{{2(x+1)}}+c $$

But the result is different. I will be thankful for any help.

Answer 1

Hint:

$$1-2x-x^2$$ factors as

$$-(x-a)(x-b)$$ where

$$a,b=1\pm\sqrt2.$$

Then you can decompose

$$\frac{b-a}{(x-a)(x-b)}=\frac1{x-b}-\frac1{x-a}$$ and integrate

$$\log|x-b|-\log|x-a|.$$

Answer 2

Hint:

$$\int\dfrac{1}{1-2x-x^2}\mathrm dx=\dfrac{-1}{\sqrt{2}}\int\dfrac{1}{\left(\frac{x+1}{\sqrt{2}}\right)^2-1}\mathrm dx$$

This is inviting for Partial Fraction Decomposition. Can you proceed?

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$$\int\dfrac{1}{2-(x+1)^2}\mathrm dx\ne\dfrac{1}{2}\int\dfrac{1}{-(x+1)^2}\mathrm dx$$

Answer 3

Use the u-substation $u=\frac{x+1}{\sqrt{2}}$ and then do a partial fraction decomposition on the fraction that you will get: $$ \int\frac{1}{1-2x-x^2}\,dx=\int\frac{1}{2-(x+1)^2}\,dx=\\ \frac{\sqrt{2}}{2}\int\frac{1}{1-\left(\frac{x+1}{\sqrt{2}}\right)^2}\frac{d}{dx}\left(\frac{x+1}{\sqrt{2}}\right)\,dx= \frac{\sqrt{2}}{2}\int\frac{1}{1-u^2}\,du=\\ \frac{\sqrt{2}}{2}\int\frac{1}{(1-u)(1+u)}\,du= \frac{\sqrt{2}}{2}\int\left(\frac{1}{2}\frac{1}{1-u}-\frac{1}{2}\frac{1}{1+u}\right)\,du=\\ \frac{\sqrt{2}}{4}\left(-\ln{|1-u|-\ln{|1+u|}}\right)+C=\\ \frac{\sqrt{2}}{4}\ln{\left|\frac{1+u}{1-u}\right|}+C. $$

And then do a back-substitution.