$42) f(x)=\dfrac{2x^4-5x^3+1}{x^2}\\ F=\displaystyle\int \left(\dfrac{2x^4-5x^3+1}{x^2}\right)dx\\ =\displaystyle\int \left(2x^2-5x+\dfrac{1}{x^2}\right)dx\\ =\dfrac{2}{3}x^3-\dfrac{5}{2}x^2-\dfrac{1}{x}+C\\ F(1)=3<=>\dfrac{2}{3}-\dfrac{5}{2}-1+C=-3=>C=\dfrac{-1}{6}\\ =>F(x)=\dfrac{2}{3}x^3-\dfrac{5}{2}x^2-\dfrac{1}{x}-\dfrac{1}{6}\\$$ 43) f(x)=cos^2x\\ F(x)=\int cos^2xdx\\ =\displaystyle\int \dfrac{1+cos2x}{2}dx\\ =\dfrac{1}{2}\left(\int dx + \int cos2xdx\right)\\ =\dfrac{1}{2}\left(x +\dfrac{1}{2} \int cos2xd(2x)\right)\\ =\dfrac{1}{2}\left(x +\dfrac{1}{2}sin2x\right)+C\\ F\left(\dfrac{\pi}{4}\right)=\pi<=>\dfrac{1}{2}\left(\dfrac{\pi}{4} +\dfrac{1}{2}\right)+C=\dfrac{-1}{4}+\dfrac{7\pi}{8}\\ =>F(x)=\dfrac{1}{2}\left(x +\dfrac{1}{2}sin2x\right)+\dfrac{-1}{4}+\dfrac{7\pi}{8}\\$  

`a)`

`f (x) = (2x^4 - 5x^3 + 1)/(x^2)`

`-> F (x) = int f(x)dx`

`-> F (x) = int (2x^4 - 5x^3 + 1)/(x^2)dx`

`-> F (x) = int (2x^2 - 5x + 1/(x^2))dx`

`-> F (x) = (2x^3)/(2 + 1) - (5x^2)/(1 + 1) + (x^{-1})/(-2 + 1) + C`

`-> F (x) = 2/(3)x^3 - 5/(2)x^2 - 1/(x) + C`

`text{Theo giả thuyết}`

`-> F (1) = -3`

`-> 2/(3) - 5/2 - 1 + C= -3`

`-> C = -1/6`

`-> F (x) = 2/(3)x^3 - 5/(2)x^2 - 1/(x) - 1/6`

`b)`

`f (x) = cos^2 x`

`-> F (x) = int cos^2 xdx`

`-> F (x) = int (1 + cos 2x)/(2)dx`

`-> F (x) = int (1/2 + (cos 2x)/(2))dx`

`-> F (x) = 1/(2)x + 1/(2).(1)/(2)sin 2x + C`

`-> F (x) = 1/2(x + sin 2x) + C`

`text{Theo giả thuyết}`

`-> F (\pi)/4 = pi`

`-> 1/(2).[(\pi)/4 + sin ((\pi)/2)] + C = pi`

`-> (\pi)/8 + 1/(4) + C = pi`

`-> C = (7\pi)/8 - 1/4`

`-> F (x) = 1/2(x + sin 2x) + (7\pi)/8 - 1/4`