Question

1.∫√x(1+x)/[√x+√（1+x）] dx 2.∫(1+x)/x(1＋xe^x) dx

Answer 1

.∫√x(1+x)/[√x+√（1+x）] dx ∫1／√x(1+x)dx
分母有理化:同√x+1 - √x
=∫(x+1)√x-x√x+1dx
=∫(x+1)√xdx-∫x√x+1dx
=∫x√xdx+∫√xdx-∫x√x+1dx
=1/(3/2+1)x^(3/2+1)+1/(1/2+1)x^(3/2)-∫x√x+1dx
=2/5x^(5/2)+2/3x^(3/2)-∫x√x+1dx

∫x√x+1dx
y=√x+1
x=yy-1
dx=2ydy
=>
∫x√x+1dx
=∫(yy-1)y*2ydy
=∫(yy-1)y*2ydy
=2∫y^4dy-2∫y^2dy
=2/5y^5-2/3y^3+C
=2/5(x+1)^5/2-2/3(x+1)^3/2+C

原式=2/5x^(5/2)+2/3x^(3/2)-[2/5(x+1)^5/2-2/3(x+1)^3/2]+C

.∫(1+x)/[x(1＋xe^x) ]dx
原式=∫(1+x)e^x/[x(e^x)(1+xe^x)]dx
=∫1/[x(e^x)(1+xe^x)]d(1+xe^x)
=∫1/xe^xdxe^x-∫1/(1+xe^x)d(1+xe^x)
=∫ln|x(e^x)/(1+xe^x)|+c