Let #I=int1/(sqrt(1+x)+sqrtx)dx#
#:. I=int1/(sqrt(1+x)+sqrtx) xx (sqrt(1+x)-sqrtx)/(sqrt(1+x)-sqrtx)dx#
#=int(sqrt(1+x)-sqrtx)/(1+x-x)dx=int(sqrt(1+x)-sqrtx)dx#
#=intsqrt(1+x)dx-intsqrtxdx=I_1-intx^(1/2) dx#
#=I_1-x^(1/2+1)/(1/2+1)=I_1-2/3x^(3/2)#, where,
#I_1=intsqrt(1+x)dx#
Let, #1+x=t^2, so, dx=2tdt#. Hence,
#I_1=intsqrt(t^2)*2tdt#
#=2int t^2dt=2*t^3/3=2/3(t^2)^(3/2)=2/3(1+x)^(3/2)#.
Altogether,
#I=I_1-2/3x^(3/2)=2/3(1+x)^(3/2)-2/3x^(3/2)#, i.e.,
#I=2/3{(1+x)^(3/2)-x^(3/2)}+C#.
In fact, we have used the substn. method because it was so
required. Otherwise, we can solve it without using substn., by the
following Rule :-
#intf(x)dx=F(x)+C rArr intf(ax+b)dx=1/a*F(ax+b)+k, a!=0#
Now, #intsqrtxdx=2/3x^(3/2)+crArrintsqrt(1+x)=1/1*2/3*(1+x)^(3/2)+k#.
Enjoy Maths.!
