Question

Integration of 1/1+v^2 -v/1+v^2 dv

Answer 1

Answer:

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Step-by-step explanation:

Let I = Integ.(2+v)/(1-v^2) dv

= integ.(2+v)/(1+v)(1-v) dv

Now let (2+v)/(1+v)(1-v) = A/(1+v) + B/(1-v)

= {A(1-v)+B(1+v)}/(1+v)(1-v)

=> 2+v = A(1-v)+B(1+v).

When v=1 gives 2+1=0+B(2) => B= 3/2

When v = -1 gives 2–1= A(1+1)=> 2A = 1 => A=1/2

So I = integ .[(1/2)/(1+v) + (3/2)/(1-v)] dv

=(1/2) log(1+v) -(3/2) log (1-v)+c