Giải thích các bước giải:
$\begin{array}{l}
b)F = \int {\dfrac{{3{x^4} - 2{x^3} + 5}}{{{x^2}}}dx} \\
= \int {\left( {3{x^2} - 2x + \dfrac{5}{{{x^2}}}} \right)dx} \\
= \int {3{x^2}dx} - \int {2xdx} + 5\int {\dfrac{1}{{{x^2}}}dx} \\
= {x^3} - {x^2} - \dfrac{5}{x} + C
\end{array}$
Ta có:
$F\left( 1 \right) = 2 \Rightarrow 1 - 1 - 5 + C = 2 \Rightarrow C = 7$
$ \Rightarrow F\left( x \right) = {x^3} - {x^2} - \dfrac{5}{x} + 7$
Vậy $F\left( x \right) = {x^3} - {x^2} - \dfrac{5}{x} + 7$
$\begin{array}{l}
c)F = \int {\dfrac{{{x^3} + 3{x^2} + 3x - 7}}{{{{\left( {x + 1} \right)}^2}}}dx} \\
= \int {\dfrac{{{{\left( {x + 1} \right)}^3} - 8}}{{{{\left( {x + 1} \right)}^2}}}dx} \\
= \int {\left( {x + 1} \right)dx} - 8\int {\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}dx} \\
= \dfrac{{{x^2}}}{2} + x + \dfrac{8}{{x + 1}} + C
\end{array}$
Ta có:
$F\left( 0 \right) = 8 \Rightarrow 0 + 0 + 8 + C = 8 \Rightarrow C = 0$
$ \Rightarrow F\left( x \right) = \dfrac{{{x^2}}}{2} + x + \dfrac{8}{{x + 1}}$
Vậy $F\left( x \right) = \dfrac{{{x^2}}}{2} + x + \dfrac{8}{{x + 1}}$
