How do you find ∫1/20arctan(1–2x2−−−−−−√)1+2x2dx?

 To evaluate the integral ∫(1/20) arctan(1-2x^2)^1+2x^2 dx, we can use the substitution u = 1-2x^2:

du/dx = -4x,

x dx = -1/4 du.

Substituting these values in the integral, we get:

∫(1/20) arctan(u) (1/2) (-1/4) du

= -(1/160) ∫arctan(u) du

= -(1/160) [u arctan(u) - ∫u/(1+u^2) du] // using integration by parts

= -(1/160) [u arctan(u) - (1/2) ln(1+u^2) + C] // where C is the constant of integration

Substituting back the value of u = 1-2x^2, we get:

∫(1/20) arctan(1-2x^2)^1+2x^2 dx

= -(1/160) [(1-2x^2) arctan(1-2x^2) - (1/2) ln(1+(1-2x^2)^2) + C]

Hence, we have obtained the antiderivative of the given function.