Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 35

int_1^2x^4(ln(x))^2dx
If f(x) and g(x) are differentiable functions, then
intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx
If we write f(x)=u and g'(x)=v, then
intuvdx=uintvdx-int(u'intvdx)dx
Using the above method of integration by parts,
intx^4(ln(x))^2dx=(ln(x))^2intx^4dx-int(d/dx(ln(x)^2)intx^4dx)dx
=(ln(x))^2*x^5/5-int(2ln(x)*1/x(x^5/5))dx
=(ln(x))^2x^5/5-2/5intx^4ln(x)dx
again applying integration by parts,
=x^5/5(ln(x))^2-2/5(ln(x)intx^4dx-int(d/dx(ln(x))intx^4dx)dx)
=x^5/5(ln(x))^2-2/5(ln(x)x^5/5-int(1/x*x^5/5)dx)
=x^5/5(ln(x))^2-2/25x^5ln(x)+2/25intx^4dx
=x^5/5(ln(x))^2-2/25x^5ln(x)+2/25*x^5/5
adding constant to the solution,
=x^5/5(ln(x))^2-2/25x^5ln(x)+2/125x^5+C
Now evaluate the definite integral,
int_1^2x^4(ln(x))^2dx=[x^5/5(ln(x))^2-2/25x^5ln(x)+2/125x^5]_1^2
=[2^5/5(ln(2))^2-2/25*2^5ln(2)+2/125(2^5)]-[1^5/5(ln(1))^2-2/25(1)^5ln(1)+2/125(1^5)]
=[32/5(ln(2))^2-64/25ln(2)+64/125]-[2/125]
=32/5(ln(2))^2-64/25ln(2)+62/125