Question #64897

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Perform the integral

#int_1^9((x-1)dx)/sqrtx=int_1^9(x/sqrtx-1/sqrtx)dx#

#=int_1^9(sqrtx-1/sqrtx)dx#

#=[2/3x^(3/2)-2sqrtx]_1^9#

#=(2/3*27-6)-(2/3-2)#

#=14-2/3#

#=40/3#

#int_1^9(x-1)/sqrtx#

Let us start by splitting the fraction into two parts:
#=int_1^9(x/sqrtx-1/sqrtx)#

Then we can write #1/sqrt x# in terms of power (#x^(-1/2#):
#=int_1^9((x^1*x^(-1/2))-x^(-1/2))#

Adding the powers in the first term, using following power rules:
#a^x*a^y=a^(x+y)#
#a^x/a^y=a^(x-y)#

#rarrint_1^9(x^(1-1/2)-x^(-1/2))#
#=int_1^9(x^(1/2)-x^(-1/2))#

Using the following power rule of integration,

We find out that :
#int_1^9(x^(1/2)-x^(-1/2))=[(x^(1/2+1))/(1/2+1)+C-x^(-1/2+1)/(-1/2+1)+C]_1^9#

#=[x^(3/2)/(3/2)+C-x^(1/2)/(1/2)+C]_1^9#
#=[2/3sqrt(x^3)+C-2sqrtx+C]_1^9#

Plotting the value of 9 then subtracting it from value of 1:
#=(2/3sqrt(9^3)-2sqrt9+C)-(2/3sqrt(1^3)-2sqrt1+C)#

Evaluating the function we get:
#=(2/3(3^3)-2(3)+C)-(2/3(1)-2(1)+C)#
#=(2(3^2)-6+C)-(2/3-2+C)#
#=(2(9)-6+C)-(2/3-6/3+C)#
#=12cancel+C-4/3cancel(-C)=36/3-(-4/3)=36/3+4/3#
#=40/3=13 1/3#

#"express the terms in the form "ax^n" and integrate"#
#"each term using the "color(blue)"power rule"#

#•color(white)(x)int(ax^n)=a/(n+1)x^(n+1)color(white)(x)n!=-1#

#int_1^9((x-1))/sqrtxdx=int_1^9((x-1))/x^(1/2)dx#

#=int_1^9(x^(1/2)-x^(-1/2))dx#

#=[2/3x^(3/2)-2x^(1/2)]_1^9#

#=(2/3xx27-6)-(2/3-2)#

#=12+4/3#

#=40/3#