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
