Evaluate the integral of 1/ ( 1 + 4x^2)
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We have to find Int [1/ (1 + 4x^2) dx].
First substitute u = 2x
=> du /dx = 2
=> du /2 = dx
Now Int [1/ (1 + 4x^2) dx]
=> Int [(1/2)*(1/ (1+u^2) du]
=> (1/2)*Int [1/ (1 + u^2) du]
Now Int [1/ (1+u^2) du] =arc tan u + C
=> (1/2)*arc tan u + C
replace u with 2x
=> (1/2)* arc tan 2x + C
Therefore the required result is
1/2)* arc tan 2x + C
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To evaluate the integral of 1/(1+4x^2).
Solution:
We know that Integral of 1/(x^2+a^2 ) = Int {1/(x^2+a^2)} dx = (1/a)arc tan (x/a) + C....(1) .
The given function 1/(4x^2+1) = (1/4){1/[x^2+(1/4)]} = (1/4){1/[x^2+(1/2)^2]}.
Therefore Int {1/(4x^2+1)} dx = (1/4) Int {1/[x^2+(1/2)^2]} dx which is like (1). Here a = 1/2.
So Int {1/(4x^2+1)} dx = (1/4)* (1/(1/2))* arc tan {x/(1/2)}+C .
Therefore Int {1/(4x^2+1)} dx = (1/2) arc tan(2x) +C.
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