The value of _ - ^ 2x(1 + x) 1 + ^2 x dx is...

Step-by-step solution

Let I = _ - ^ 2x(1 + x) 1 + ^2 x dx Splitting the integral, we get: I = _ - ^ 2x 1 + ^2 x dx + _ - ^ 2x x 1 + ^2 x dx The first integral is zero because the integrand f(x) = 2x 1 + ^2 x is an odd function. The second integral has an even integrand, so: I = 2 ₀^ 2x x 1 + ^2 x dx = 4 ₀^ x x 1 + ^2 x dx Using the property ₀^ a f(x)dx = ₀^ a f(a-x)dx , we get: I = 4 ₀^ ( - x) ( - x) 1 + ^2( - x) dx = 4 ₀^ ( - x) x 1 + ^2 x dx Adding the two expressions for I : 2I = 4 ₀^ x 1 + ^2 x dx I = 2 ₀^ x 1 + ^2 x dx Substituting