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Question

The value of the integral
1201+3((x+1)2(1x)6)14 dx  
is


Solution

I=1201+3((x+1)2(1x)6)14 dx=1201+3(1x)1x2dx 
Let x=sinθdx=cosθ dθ
I=π60(1+3)cosθcosθ(1sinθ)dθ=(1+3)π60(1+sinθ)cos2θdθ
I=(1+3)π60sec2θ dθ+π60secθtanθ dθ
I=(1+3)(tanθπ60+secθπ60)
I=(1+3)(13+(231))
I=(3+1)(31)=((3)212)=2

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