Integrate (t+1)/(2(1+t^2)) wrt t, limits: [0,1]?

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( t + 1 ) / [ 2 ( 1 + t ² ) ]

t / [ 2 ( 1 + t ² ) ] + 1 / [ 2 ( 1 + t ² ) ]

I = (1/4) ∫ 2t dt / ( 1 + t ² ) + (1 /2 ) ∫ [ dt / ( 1 ² + t ² ) ]

I = (1/4) log ( 1 + t ² ) + (1/2) tan^(-1) t

Inserting limits , this becomes :-

I = (1/4) log 2 + (1/2) tan^(-1) 1

I = (1/4) log 2 + (1/2) ( π/4)

I = (1/4) log 2 + π / 8...Show more