Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 32

int_1^2(ln(x))^2/x^3dx
If f(x) and g(x) are differentiable function, then
intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx
If we rewrite f(x)=u and g'(x)=v, them
intuvdx=uintvdx-int(u'intvdx)dx
Using the above method of integration by parts,
Let's first evaluate the indefinite integral,
int(ln(x))^2/x^3dx=(ln(x))^2int1/x^3dx-int(d/dx(ln(x))^2int(1/x^3)dx)dx
=(ln(x))^2(x^(-3+1)/(-3+1))-int(2ln(x)(1/x)(x^(-3+1)/(-3+1)))dx
=(ln(x))^2(x^(-2)/-2)-int(2ln(x)(1/x)(x^(-2))/-2)dx
=-1/(2x^2)(ln(x))^2+int(ln(x)/x^3)dx
again applying integration by parts
=-1/(2x^2)(ln(x))^2+ln(x)*int(1/x^3)dx-int(d/dx(ln(x)int(1/x^3)dx)dx
=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-3+1)/(-3+1))-int(1/x)((x^-3+1)/(-3+1))dx
=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-2)/-2)-int(1/(-2x^3))dx
=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2int(1/x^3)dx
=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2(x^(-3+1)/(-3+1))
=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)-1/(4x^2)
add a constant C to the solution,
=-1/(2x^2)((ln(x))^2+ln(x)+1/2)+C
Now let's evaluate definite integral,
int_1^2(ln(x))^2/x^3dx=[-1/(2x^2)((ln(x))^2+ln(x)+1/2)]_1^2
=[-1/(2*2^2)((ln(2))^2+ln(2)+1/2)]-[-1/(2*1^2)((ln(1))^2+ln(1)+1/2)]
=[-1/8(ln(2))^2-ln(2)/8-1/16+1/4]
=3/16-(ln(2))^2/8-ln(2)/8