Bài 41
a) \(y = 2x\left( {1 – {x^{ – 3}}} \right);\) b) \(y = 8x – {2 \over {{x^{{1 \over 4}}}}};\)
c) \(y = {x^{{1 \over 2}}}\sin \left( {{x^{{3 \over 2}}} + 1} \right);\) d) \(y = {{\sin \left( {2x + 1} \right)} \over {{{\cos }^2}\left( {2x + 1} \right)}};\)
a) \(\int {2x\left( {1 – {x^{ – 3}}} \right)} dx = \int {\left( {2x – 2{x^{ – 2}}} \right)dx = {x^2} + {2 \over x} + C} \)
b) \(\int {\left( {8x – {2 \over {{x^{{1 \over 4}}}}}} \right)dx = } \int {\left( {8x – 2{x^{ – {1 \over 4}}}} \right)} dx = 4{x^2} – {8 \over 3}{x^{{3 \over 4}}} + C\)
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c) Đặt
\(\eqalign{
& u = {x^{{3 \over 2}}} + 1 \Rightarrow du = {3 \over 2}{x^{{1 \over 2}}}dx \Rightarrow {x^{{1 \over 2}}}dx = {2 \over 3}du \cr
& \int {{x^{{1 \over 2}}}\sin\left( {{x^{{3 \over 2}}} + 1} \right)dx = {2 \over 3}\int {\sin udu = – {2 \over 3}\cos u + C = – {2 \over 3}\cos \left( {{x^{{3 \over 2}}} + 1} \right)} + C} \cr} \)
d) Đặt \(u = \cos \left( {2x + 1} \right) \Rightarrow du = – 2\sin \left( {2x + 1} \right)dx \Rightarrow \sin \left( {2x + 1} \right)dx = – {1 \over 2}du\)
Do đó \(\int {{{\sin \left( {2x + 1} \right)} \over {{{\cos }^2}\left( {2x + 1} \right)}}} dx = – {1 \over 2}\int {{{du} \over {{u^2}}} = {1 \over {2u}} + C = {1 \over {2\cos \left( {2x + 1} \right)}}} + C\)