Answer :

Let us assume

Adding – 1 and + 1

–

Let

Thus I = I_{1} – I_{2} …….equation 1

Solving for I_{1}

since

I_{1} = [tan ^{– 1}(∞) – tan ^{– 1}(0)]

I_{1} = π/2 ……….equation 2

Solving for I_{2}

Let .....…..equation 3

a + b = 0; a + c = 1; b + c = 0

solving we get

a = c = 1/2

b = – 1/2

substituting the values in equation 3

Thus substituting the values in I_{2}, thus

Solving :

Let 1 + x^{2} = y

2xdx = dy

For x = ∞

y = ∞

For x = 0

y = 0

substituting values

Thus

……….equation 4

Substituting values equation 2 and equation 4 in equation 1

Thus

I = I_{1 –} I_{2}

I = π/2 – π/4

I = π/4

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