Question

求ln(1+x)/√xdx的不定积分 还有一题xarctanxdx的不定积分

Answer 1

∫ ln(1+x)/√x dx
= 2∫ ln(1+x)/(2√x) dx
= 2∫ ln(1+x) d√x
= 2ln(1+x) * √x - 2∫ √x dln(1+x)，integration by part
= 2(√x)ln(1+x) - 2∫ √x/(1+x) dx
= 2(√x)ln(1+x) - 2∫ (2√x*√x)/[2√x*(1+x)] dx
= 2(√x)ln(1+x) - 4∫ [1+(√x)²-1]/[1+(√x)²] d√x
= 2(√x)ln(1+x) - 4∫ d√x + 4∫ d√x/[1+(√x)²]
= 2(√x)ln(1+x) - 4√x + 4arctan(√x) + C

∫ xarctanx dx
= ∫ arctanx d(x²/2)
= (1/2)x²arctanx - (1/2)∫ x² d(arctanx)，integration by part
= (1/2)x²arctanx - (1/2)∫ x²/(1+x²) dx
= (1/2)x²arctanx - (1/2)∫ (1+x²-1)/(1+x²) dx
= (1/2)x²arctanx - (1/2)∫ dx + (1/2)∫ dx/(1+x²)
= (1/2)x²arctanx - x/2 + (1/2)arctanx + C