Question

integrate 1/(1-x-x^2) dx​

Answer 1

Answer:

I=integ. of dx/(1-x-x^2)

I=integ. of dx/[1 -(x^2+x)]

I=integ. of dx/[ 1 - (x^2 +x +1/4)+1/4]

I=integ. of dx/[{(5^1/2)/2} ^2-(x+1/2)^2]

Let x+1/2= { (5)^1/2}/2 sinA

dx={(5)^1/2} cosA.dA

I=integ.of 2/(5)^1/2 secA dA

I=2/(5)^1/2 log(secA +tanA)+logk

With the help sinA=(2x+1)/(5)^1/2 put secA={(5)^1/2}/{2(1-x-x^2)^1/2} and tanA=(2x+1)/{2(1-x-x^2)^1/2}.

I=2/(5)^1/2log[(2x+1+5^1/2)/[2(1-x-x^2)^1/2]+log k

I=2/(5)^1/2.log[(2x+1+5^1/2)/(1-x-x^2)^1/2]+[logk- 2/(5)^1/2. log 2]

I=2/(5)^1/2.log[(2x+1+5^1/2)/(1-x-x^2)^1/2]+logc

Step-by-step explanation:

hope it's helpful