高等数学，求不定积分

Question

希望有详细过程

Answer 1

一、
√(1 + x²) = √(1 + tan²y) = √sec²y = secy
设x = tany，dx = sec²y dy
∫ x⁵/√(1 + x²) dx
= ∫ tan⁵ysec²y/secy dy
= ∫ tan⁵ysecy dy
= ∫ (sec²y - 1)² dsecy
= ∫ (sec⁴y - 2sec²y + 1) dsecy
= (1/5)sec⁵y - (2/3)sec³y + secy + C
= (1/5)(1 + x²)^(5/2) - (2/3)(1 + x²)^(3/2) + √(1 + x²) + C
= (1/5)(3x⁴- 4x² + 8)√(1 + x²) + C
二、
令u² = e^x - 1，2u du = e^x dx
∫ xe^x/√(e^x - 1) dx
= 2∫ ln(u² + 1) du
= 2uln(u² + 1) - 4∫ u²/(u² + 1) du
= 2uln(u² + 1) - 4∫ (u² + 1 - 1)/(u² + 1) du
= 2uln(u² + 1) - 4∫ du + 4∫ du/(u² + 1)
= 2uln(u² + 1) - 4u + 4arctan(u) + C
= 2x√(e^x - 1) - 4√(e^x - 1) + 4arctan√(e^x - 1) + C
= 2(x - 2)√(x - 2) + 4arctan√(e^x - 1) + C
三、
令t² = x + 1，2t dt = dx
∫ lnx/√(1 + x) dx
= 2∫ ln(t² - 1) dt
= 2tln(t² - 1) - 2∫ t d[ln(t² - 1)]
= 2tln(t² - 1) - 4∫ t²/(t² - 1) dt
= 2tln(t² - 1) - 4∫ (t² - 1 + 1)/(t² - 1) dt
= 2tln(t² - 1) - 4∫ dt - 4∫ dt/[(t + 1)(t - 1)]
= 2tln(t² - 1) - 4t - 2∫ [(t + 1) - (t - 1)]/[(t + 1)(t - 1)] dt
= 2tln(t² - 1) - 4t - 2∫ [1/(t - 1) - 1/(t + 1)] dt
= 2tln(t² - 1) - 4t + 2ln|(t + 1)/(t - 1)| + C
= 2√(x + 1)ln|x| - 4√(x + 1) + 2ln| [√(x + 1) + 1]/[√(x + 1) - 1] | + C
= 2(ln|x| - 2)√(x + 1) + 4arccoth√(x + 1) + C