Question

有大神会这两道不定积分吗?

Answer 1

乱七八糟答案真多，详细过程如图rt所示……希望能帮到你解决你心中的问题

Answer 2

∫ 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

追问

我为啥看不懂啊