Question

这个不定积分怎么求？

exp(arctan(x))/(1+x^2)^3/2这个函数的不定积分

Answer 1

楼上的解法用的是三角代换，换来换去的麻烦，但本题可以直接用分部积分:
∫e^(arctanx)/[(1+x^2)^3/2]dx
=∫1/√(1+x^2)d[e^(arctanx)]
=e^(arctanx)/√(1+x^2)+∫x*e^(arctanx)/[(1+x^2)^3/2]dx
=e^(arctanx)/√(1+x^2)+∫x/√(1+x^2)d[e^(arctanx)]
=e^(arctanx)/√(1+x^2)+x*e^(arctanx)/√(1+x^2)-∫e^(arctanx)/[(1+x^2)^3/2]dx
所以2∫e^(arctanx)/[(1+x^2)^3/2]dx=e^(arctanx)/√(1+x^2)+x*e^(arctanx)/√(1+x^2)
即：∫e^(arctanx)/[(1+x^2)^3/2]dx=[(x+1)e^(arctanx)]/[2√(1+x^2)]+C

Answer 2

令t = arctanx，x = tant，dx = sec²t dt
sint = x/√(1 + x²)，cost = 1/√(1 + x²)

∫ [e^(arctanx)]/(1 + x²)^(3/2) dx
= ∫ (e^t/sec³t)(sec²t) dt
= ∫ e^t * cost dt
= ∫ e^t d(sint)
= e^t * sint - ∫ e^t * sint dt
= e^t * sint + ∫ e^t d(cost)
= e^t * sint + e^t * cost - ∫ e^t * cost dt
2∫ e^t * cost dt = (sint + cost) * e^t
∫ e^t * cost dt = (1/2)(sint + cost) * e^t + C
==> ∫ [e^(arctanx)]/(1 + x²)^(3/2) dx = (1/2)[x/√(1 + x²) + 1/√(1 + x²)] * e^(arctanx) + C
= [(x + 1)e^(arctanx)]/[2√(1 + x²)] + C