∫﹙x²＋√x³＋3﹚／﹙√x﹚dx

∫﹙x²＋√x³＋3﹚／﹙√x﹚dx
∫﹙x²＋√x³＋3﹚／﹙√x﹚dx

∫﹙x²＋√x³＋3﹚／﹙√x﹚dx
d(√x)=1/(2√x) dx => 2d(√x)=1/√x dx
则
∫(x²＋√x³＋3)/(√x)dx
=2∫﹙x²＋√x³＋3﹚d(√x)
=2(√x^5/5+x²/4+3√x)+C
=(2/5)√x^5+x²/2+6√x+C

dx是在外面吗?如果是,那么拆项就行了,(x²＋√x³＋3﹚／﹙√x﹚就变成∫(X^3/2+x+3X^-1/2)dx,然后简单积分就行了

∫﹙x²＋√x³＋3﹚／﹙√x﹚dx
=∫[x^(3/2)＋x＋3／﹙√x﹚]dx
=2/5x^(5/2)+x^2/2+6√x+C