Bài 41: Tìm nguyên hàm của các hàm số sau:
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) \(\int {\left( {8x – {2 \over {{x^{{1 \over 4}}}}}} \right)dx = } \int {\left( {8x – 2{x^{ – {1 \over 4}}}} \right)} dx\)
\(= 4{x^2} – {8 \over 3}{x^{{3 \over 4}}} + C\)
c) Đặt
\(\eqalign{
& u = {x^{{3 \over 2}}} + 1 \Rightarrow du = {3 \over 2}{x^{{1 \over 2}}}dx \Rightarrow {x^{{1 \over 2}}}dx = {2 \over 3}du \cr
& \int {{x^{{1 \over 2}}}\sin\left( {{x^{{3 \over 2}}} + 1} \right)dx = {2 \over 3}\int {\sin udu} }\cr&= – {2 \over 3}\cos u + C = – {2 \over 3}\cos \left( {{x^{{3 \over 2}}} + 1} \right) + C\cr} \)
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 \(\int {{{\sin \left( {2x + 1} \right)} \over {{{\cos }^2}\left( {2x + 1} \right)}}} dx = – {1 \over 2}\int {{{du} \over {{u^2}}} = {1 \over {2u}}} + C\)
\(= {1 \over {2\cos \left( {2x + 1} \right)}} + C\)
Bài 42: a) \(y = {1 \over {{x^2}}}\cos \left( {{1 \over x} – 1} \right)\); b) \(y = {x^3}{\left( {1 + {x^4}} \right)^3}\);
c) \(y = {{x{e^{2x}}} \over 3}\); d) \(y = {x^2}{e^x}\).
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a) Đặt \(u = {1 \over x} – 1 \Rightarrow du = – {1 \over {{x^2}}}dx \Rightarrow {{dx} \over {{x^2}}} = – du\)
Do đó \(\int {{1 \over {{x^2}}}} \cos \left( {{1 \over x} – 1} \right)dx = – \int {\cos udu }\)
\(= – \sin u + C = – \sin \left( {{1 \over x} – 1} \right) + C\)
b) Đặt \(u = 1 + {x^4} \Rightarrow du = 4{x^3}dx \Rightarrow {x^3}dx = {{du} \over 4}\)
\(\int {{x^3}{{\left( {1 + {x^4}} \right)}^3}dx = {1 \over 4}\int {{u^3}du = {{{u^4}} \over {16}} + C} } \)
\(= {1 \over {16}} {\left( {1 + {x^4}} \right)^4} + C\)
c) Đặt
\(\left\{ \matrix{
u = {x \over 3} \hfill \cr
dv = {e^{2x}}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = {1 \over 3}dx \hfill \cr
v = {1 \over 2}{e^{2x}} \hfill \cr} \right.\)
Suy ra: \(\int {{{x{e^{2x}}} \over 3}dx = {1 \over 6}x{e^{2x}} – {1 \over 6}\int {{e^{2x}}dx} } \)
\(= {1 \over 6}x{e^{2x}} – {1 \over {12}}{e^{2x}} + C \)
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d) Đặt
\(\left\{ \matrix{
u = {x^2} \hfill \cr
dv = {e^x}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = 2xdx \hfill \cr
v = {e^x} \hfill \cr} \right.\)
Suy ra \(\int {{x^2}{e^x}dx = {x^2}{e^x} – 2\int {x{e^x}dx} } \) (1)
Đặt
\(\left\{ \matrix{
u = x \hfill \cr
dv = {e^x}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = {e^x} \hfill \cr} \right.\)
Do đó: \(\int {x{e^x}dx = x{e^x} – \int {{e^x}dx = x{e^x} – {e^x} + C} } \)
Từ (1) suy ra \(\int {{x^2}{e^x}dx = {x^2}{e^x} – 2x{e^x} + 2{e^x} + C} \)
\(= {e^x}\left( {{x^2} – 2x + 2} \right) + C\)
Bài 43: a) \(y = x{e^{ – x}}\); b) \(y = {{\ln x} \over x}\).
a) Đặt
\(\left\{ \matrix{
u = x \hfill \cr
dv = {e^{ – x}}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = – {e^{ – x}} \hfill \cr} \right.\)
Suy ra \(\int {x{e^{ – x}}dx = – x{e^{ – x}} + \int {{e^{ – x}}dx} }\)
\(= – x{e^{ – x}} – {e^{ – x}} + C = – {e^{ – x}}\left( {x + 1} \right) + C \)
b) Đặt \(u = \ln x \Rightarrow du = {{dx} \over x}\)
Do đó \(\int {{{\ln x} \over x}} dx = \int {udu = {{{u^2}} \over 2}} + C = {{{{(\ln x)}^2}} \over 2} + C\)
Bài 44: Tìm hàm số \(y = f(x)\) nếu biết \(dy = 12x{\left( {3{x^2} – 1} \right)^3}dx\) và \(f(1) = 3\).
Ta có \(y = f\left( x \right) = \int {dy = 12\int {x{{\left( {3{x^2} – 1} \right)}^3}dx} } \)
Đặt \(u = 3{x^2} – 1 \Rightarrow du = 6xdx \Rightarrow xdx = {{du} \over 6}\)
Do đó \(f\left( x \right) = 2\int {{u^3}} du = {{{u^4}} \over 2} + C = {1 \over 2}{\left( {3{x^2} – 1} \right)^4} + C\)
Vì \(f\left( 1 \right) = 3\) nên \({1 \over 2}{2^4} + C = 3 \Rightarrow C = – 5\)
Vậy \(f\left( x \right) = {1 \over 2}{\left( {3{x^2} – 1} \right)^4} – 5\)
