Let u = tan⁻¹x, then du/dx = 1/(1+x²).
Using the formula for the derivative of inverse tangent, we have:
tan(u) = x
sec²(u) du/dx = 1
du/dx = cos²(u)
Substituting into the original expression, we get:
∫(tan⁻¹x × e^tan⁻¹x)/(1+x²) dx = ∫(tan⁻¹x × e^u × cos²(u)) du
Using integration by parts with u = tan⁻¹x and dv = e^u × cos²(u) du, we get:
v = (1/2) e^u (sin(u) + u cos(u))
∫(tan⁻¹x × e^u × cos²(u)) du = (1/2) e^u (sin(u) + u cos(u)) tan⁻¹x - ∫[(sin(u) + u cos(u)) / (1+x²)] dx
Substituting back u = tan⁻¹x, we get:
∫(tan⁻¹x × e^tan⁻¹x × cos²(tan⁻¹x)) dx = (1/2) e^tan⁻¹x (x sin(tan⁻¹x) + cos(tan⁻¹x)) tan⁻¹x - ∫[(x cos(tan⁻¹x) + sin(tan⁻¹x)) / (1+x²)] dx
Using the identity sin(tan⁻¹x) = x / √(1+x²) and cos(tan⁻¹x) = 1 / √(1+x²), we simplify the expression to:
∫(tan⁻¹x × e^tan⁻¹x × cos²(tan⁻¹x)) dx = (1/2) x e^tan⁻¹x + (1/2) ∫[e^tan⁻¹x / (1+x²)] dx
The remaining integral can be solved using another substitution with v = tan⁻¹x, which results in:
∫(tan⁻¹x × e^tan⁻¹x × cos²(tan⁻¹x)) dx = (1/2) x e^tan⁻¹x + (1/2) ln(1+x²) + C, where C is the constant of integration