Question

integration of 1/1+4x^2​

Answer 1

Let tanA=2x, so sec^2AdA=2dx and 4x^2=tan^2A. 1+tan^2A=(cos^2A+sin^2A)/cos^2A=sec^2A.

1/(1+4x^2)=1/(1+tan^2A)=cos^2A. dx=sec^2AdA/2.

dx/(1+4x^2)=(sec^2AdA/2)cos^2A=dA/2. Integral of dA/2 is A=tan^-1(2x)/2.

Applying limits: (1/2)(tan^-1(1)-tan^-1(0))=(pi)/8

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Answer 2

hope it helps..........