This explanation is a bit long, but I couldn't find a quicker way to do it...
The integral is a linear application, so you can already split the function under the integral sign.
#int_1^4 (x^4 - x^3 + (sqrt(x-1)/x^2))dx# = #int_1^4 x^4dx - int_1^4x^3dx + int_1^4sqrt(x-1)/x^2dx#
The 2 first terms are polynomial functions, so they're easy to integrate. I show you how to do it with #x^4#.
#intx^4dx = x^5/5# so #int_1^4x^4dx = 4^5/5 - 1/5 = 1023/5#. You do the exact same thing for #x^3#, the result is #255/4#.
Finding #intsqrt(x-1)/x^2dx# is a bit long and complicated. First you multiply the fraction by #sqrt(x-1)/sqrt(x-1)# and then you change the variable : let's say #u = sqrt(x-1)#. So #du=1/(2sqrt(x-1))dx# and you now have to find #2intu^2/(u^2 + 1)^2du#. In order to find it, you need the partial fraction decomposition of the rational function #x^2/(x^2 + 1)^2#.
#x^2/(x^2 + 1)^2 = (ax+b)/(x^2 +1) + (cx+d)/(x^2+1)^2# with #a,b,c,d in RR#. After calculus, we find out that #x^2/(x^2 + 1)^2 = 1/(x^2 +1) - 1/(x^2+1)^2#, which means that #2intu^2/(u^2 + 1)^2du = 2(int(du)/(u^2+1) - int(du)/(u^2+1)^2)#
#int(du)/(u^2+1)^2# is well known, it is #arctan(u)/2 + u/(2(1+u^2))#.
Finally, #2intu^2/(u^2 + 1)^2du = 2(arctan(u) - arctan(u)/2 - u/(2(1+u^2))) = arctan(u) - u/(1+u^2)#
You replace #u# by its original expression with #x# to have #intsqrt(x-1)/x^2dx#, which is #arctan(sqrt(x-1)) - sqrt(x-1)/x#
So finally, #int_1^4sqrt(x-1)/x^2dx = arctan(sqrt3) - sqrt3/4#
