$\int \frac{1 + V}{1 - 2 V - V^{2}} d V =$

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Solution

$I = \int \frac{1 + V}{1 - 2 V - V^{2}} d V$

$I = \int \frac{1 + V}{- (V^{2} + 2 V - 1)} d V$

Let
$t = V^{2} + 2 V - 1$

$\frac{d t}{d V} = 2 V + 2$

$\frac{d t}{2} = (V + 1) d V$

Therefore,
$I = - \frac{1}{2} \int \frac{1}{t} d t$

$I = - \frac{1}{2} ln (t) + C$

On putting the value of $t$ , we get
$I = - \frac{1}{2} ln (V^{2} + 2 V - 1) + C$

Hence, this is the answer.

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