Question

integration of 1+v/1-2v-v^2​

Answer 1

Answer:

in integration the units of the physical quantities get changers

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Answer 2

Let

$I \: = \displaystyle\int\limits \frac{1 + v}{1 - 2v - {v}^{2} } dv \\ \\ = \displaystyle\int\limits \frac{1 + v}{2 - 1 - 2v - {v}^{2} } dv \\ \\ = \displaystyle\int\limits \frac{ 1+ v}{2 - (v + 1) {}^{2} } dv \\ \\ = \displaystyle\int\limits \frac{2(1 + v)}{2( \sqrt{2} - v - 1)( \sqrt{2} + v + 1) } dv \\ \\ = \displaystyle\int\limits \frac{1}{2( \sqrt{2} - v - 1) } dv - \displaystyle\int\limits \frac{1}{2( \sqrt{2} + v + 1)} dv \\ \\ = - \frac{\texttt{ In}( \sqrt{2} - v - 1)}{2} - \frac{\texttt{ In}( \sqrt{2} + v + 1) }{2} \\ \\ = - \frac{\texttt{ In} (2 - (v + 1) {}^{2} ) }{2} + C$

(C is integral constant )

____________________________

Let,

$1 - 2v - {v}^{2} = z \\ \\ \implies (- 2 - 2v)dv = dz \\ or \\ \implies \: - 2(1 + v)dv = dz \\ \\ \implies \displaystyle\int\limits \: \left(- \frac {1}{2} \right) \frac{dz}{z} = - \frac{1}{2} \texttt{In}|z| + C \\ \\ \implies \: \left( - \frac{1}{2} \right) \texttt{In} \: |1 - 2v - {v}^{2}| + C$

(C is integral constant)