Câu 3.3 trang 141 Sách BT Giải Tích 12 nâng cao: Tìm...

Tìm

a) $\int {{{{x^2} - 3x} \over x}} dx$                  b) $\int {{{4{x^3} + 5x - 1} \over {{x^2}}}} dx$

c) $\int {{{{{\left( {x + 2} \right)}^2}} \over {{x^4}}}} dx$                  d) $\int {{{{{\left( {{x^2} + 1} \right)}^2}} \over {{x^2}}}} dx$

Giải

a) $\int {{{{x^2} - 3x} \over x}} dx= \int {(x - 3)} dx={{{x^2}} \over 2} - 3x + C$                    

b) $\int {{{4{x^3} + 5x - 1} \over {{x^2}}}} dx= \int {\left( {4x + {5 \over x} - {1 \over {{x^2}}}} \right)} dx$

$=2{x^2} + 2x + {1 \over x} + C$

c) $\int {{{{{\left( {x + 2} \right)}^2}} \over {{x^4}}}} dx = \int {{{{x^2} + 4x + 4} \over {{x^4}}}}  = \int {\left( {{1 \over {{x^2}}} + {4 \over {{x^3}}} + {4 \over {{x^4}}}} \right)} dx$

$=- {1 \over x} - {2 \over {{x^2}}} - {4 \over {3{x^3}}} + C$    

d) $\int {{{{{\left( {{x^2} + 1} \right)}^2}} \over {{x^2}}}} dx = \int {{{{x^4} + 2{x^2} + 1} \over {{x^4}}}}$

$= \int {\left( {1 + {2 \over {{x^2}}} + {1 \over {{x^4}}}} \right)} dx={{{x^3}} \over 3} + {{{x^2}} \over 2} - {1 \over x} + C$