To find the area between #f(x)# and x-axis is to find the definite integral of this function between #[1,2]#
#int_1^2f(x)#
#=int_1^2x^2-ln(x^2+1)dx#
#=color(brown)(int_1^2x^2dx)-color(blue)(int_1^2ln(x^2+1)dx)#
Let us compute the first integral #color(brown)(int_1^2x^2dx#
As we know integral of the polynomial #intx^n=x^(n+1)/(n+1)#
So #color(brown)(intx^2dx=x^3/3)#
#color(brown)(int_1^2x^2dx=2^3/3-1^3/3=8/3-1/3=7/3#
Now, Let us compute the second integral #color(blue)(int_1^2ln(x^2+1)dx)# using integration by parts .
Let
#color(red)(u(x)=lnx^2+1)rArrdcolor(|red)(u(x)=(2x)/(x^2+1)dx)#
then #color(purple)(dv=dx)rArrcolor(purple)(v(x)=x)#
#color(blue)(intln(x^2+1)dx)#
#=color(red)(u(x))*color(purple)(v(x))-intcolor(purple)(v(x))*color(red)(du(x))#
#intcolor(purple)(v(x))*color(red)(du(x))#
#=int x*(2x)/(x^2+1)dx#
#=int (2x^2)/(x^2+1)dx#
#=2int(x^2)/(x^2+1)dx#
#=2int1-1/(x^2+1)dx#
#=2intdx-2int1/(x^2+1)dx#
#=color(orange)(2x-2arctanx)#
So,
#color(blue)(intln(x^2+1)dx)# #=color(red)(u(x))*color(purple)(v(x))-intcolor(purple)(v(x))*color(red)(du(x))#
#=color(purple)xcolor(red)(ln(x^2+1))-(color(orange)(2x-2arctanx))#
#=xln(x^2+1)-2x+2arctanx#
#color(blue)(int_1^2ln(x^2+1)dx)#
#=2ln(2^2+1)-2*2+2arctan2-(1*ln(1^2+1)-2*1+2arctan1)#
#=2ln5-4+2arctan2-(ln2-2+2arctan1)#
#=2ln5-4+2arctan2-ln2+2-2*pi/4#
#=2ln5-ln2-4+2+2arctan2-pi/2#
#=ln(5^2)-ln2-2+2arctan2-pi/2#
#color(blue)(=ln(5^2/2)-2+2arctan2-pi/2#
#int_1^2f(x)#
#=int_1^2x^2-ln(x^2+1)dx#
#=color(brown)(int_1^2x^2dx)-color(blue)(int_1^2ln(x^2+1)dx)#
#=color(brown)(7/3)-color(blue)(ln(5^2/2)-2+2arctan2-pi/2#
#=color(brown)(7/3)-color(blue)(ln(5^2/2)-2+2arctan2-pi/2#
#=7/3-2-ln(5^2/2)+2arctan2-pi/2#
#=7/3-6/3-ln(5^2/2)+2arctan2-pi/2#
#=1/3-ln(5^2/2)+2arctan2-pi/2#
