令u = √√x ,则x = u^4 ,dx = 4u³du
∫√(1+√x)/(x^(7/4)) dx
=∫√(1+u²)/u^7 *4u³du
=∫4√(1+u²)/u^4 du
令u = tan t ,则1+u² = sec²t ,du = sec²tdt
∫4√(1+u²)/u^4 du
=∫4sec³t/(tant)^4 dt
=∫4cost/(sint)^4 dt
=4∫dsint/(sint)^4
=-4/(3sin³t) + C
用t = arctan u和u = √√x反代入上式
得到原式 = -4(1+u²)√(1+u²)/3u³ + C = -4(1+√x)^(3/2) / 3x^(3/4) + C