Question

有谁会做这道定积分的题啊

Answer 1

以上，请采纳。

Answer 2

如下图所示

Answer 3

∫ tan⁻¹x/[x²(1 + x²)] dx
= ∫ tan⁻¹x d(- 1/x - tan⁻¹x)
= tan⁻¹x · (- 1/x - tan⁻¹x) - ∫ (- 1/x - tan⁻¹x) d(tan⁻¹x)
= - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ (1/x + tan⁻¹x)/(1 + x²) dx
= - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ [(1 + x²) - x²]/[x(1 + x²)] + ∫ tan⁻¹x/(1 + x²) dx
= - (tan⁻¹x)/x - (tan⁻¹x)² + ∫ 1/x dx - ∫ x/(1 + x²) dx + ∫ tan⁻¹x d(tan⁻¹x)
= - (tan⁻¹x)/x - (tan⁻¹x)² + ln|x| - (1/2)ln(1 + x²) + (1/2)(tan⁻¹x)² + C
= - (1/2)ln(1 + x²) - (1/2)(tan⁻¹x)² - (tan⁻¹x)/x + ln|x| + C