可以再问一个吗x(arctanx)^2的不定积分

1个回答

  • ∫ x(arctanx)² dx

    = ∫ (arctanx)² d(x²/2)

    = (1/2)x²(arctanx)² - (1/2)∫ x² * 2(arctanx) * 1/(1 + x²) dx,分部积分法

    = (1/2)x²(arctanx)² - ∫ x²/(1 + x²) * arctanx dx

    = (1/2)x²(arctanx)² - ∫ [(1 + x²) - 1]/(1 + x²) * arctanx dx

    = (1/2)x²(arctanx)² - ∫ [1 - 1/(1 + x²)] * arctanx dx

    = (1/2)x²(arctanx)² - ∫ arctanx dx + ∫ arctanx/(1 + x²) dx,中间那个用分部积分法

    = (1/2)x²(arctanx)² + ∫ arctanx d(arctanx) - [x * arctanx - ∫ x d(arctanx)]

    = (1/2)x²(arctanx)² + (1/2)(arctanx)² - xarctanx + ∫ x/(1 + x²) dx

    = (1/2)x²(arctanx)² + (1/2)(arctanx)² - xarctanx + (1/2)∫ 1/(1 + x²) d(x² + 1)

    = (1/2)x²(arctanx)² + (1/2)(arctanx)² - xarctanx + (1/2)ln(1 + x²) + C