Question #4d43b

1 Answer
Aug 17, 2016

2/3{(1+x)^(3/2)-x^(3/2)}+C23{(1+x)32x32}+C.

Explanation:

Let I=int1/(sqrt(1+x)+sqrtx)dxI=11+x+xdx

:. 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.!