(\int {{{4x - 1} \over {4{x^2} - 2x + 5}}dx} \) Bằng?

\(\int {{{4x – 1} \over {4{x^2} – 2x + 5}}dx} \) bằng?

A. \({1 \over 2}\ln \left( {4{x^2} – 2x + 5} \right) + C\)

B. \( – {1 \over {4{x^2} – 2x + 5}} + C\)

C. \({1 \over {4{x^2} – 2x + 5}} + C\)

D. \( – \ln \left( {4{x^2} – 2x + 5} \right) + C\)

Hướng dẫn

Chọn đáp án là A

Lời giải chi tiết:

Hướng dẫn

giải chi tiết

\(F\left( x \right) = \int {{{4x – 1} \over {4{x^2} – 2x + 5}}dx} \)

Ta có: \(4{x^2} – 2x + 5 = {\left( {2x} \right)^2} – 2.\left( {2x} \right).{1 \over 2} + {1 \over 4} + {{19} \over 4} = {\left( {2x – {1 \over 2}} \right)^2} + {{19} \over 4}\) \( \Rightarrow F\left( x \right) = \int {{{2\left( {2x – {1 \over 2}} \right)} \over {{{\left( {2x – {1 \over 2}} \right)}^2} + {{19} \over 4}}}dx} \)

Đặt \(2x – {1 \over 2} = {{\sqrt {19} } \over 2}\tan t \Rightarrow 2dx = {{\sqrt {19} } \over 2}{{dt} \over {{{\cos }^2}t}} = {{\sqrt {19} } \over 2}\left( {1 + {{\tan }^2}t} \right)dt\)

\(\eqalign{ & \Rightarrow F\left( t \right) = \int {{{{{\sqrt {19} } \over 2}\tan t} \over {{{19} \over 4}\left( {{{\tan }^2}t + 1} \right)}}{{\sqrt {19} } \over 2}\left( {1 + {{\tan }^2}t} \right)dt} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\tan tdt} = \int {{{\sin t} \over {\cos t}}dt} = – \int {{{d\left( {\cos t} \right)} \over {\cos t}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – \ln \left| {\cos t} \right| + C\,\,\,\left( {C = const} \right) \cr} \)

Ta có:

\(\eqalign{ & 2x – {1 \over 2} = {{\sqrt {19} } \over 2}\tan t \Rightarrow \tan t = {2 \over {\sqrt {19} }}\left( {2x – {1 \over 2}} \right) \Rightarrow {\tan ^2}t + 1 = {4 \over {19}}{\left( {2x – {1 \over 2}} \right)^2} + 1 = {1 \over {{{\cos }^2}t}} \cr & \Rightarrow {\cos ^2}t = {1 \over {{4 \over {19}}{{\left( {2x – {1 \over 2}} \right)}^2} + 1}} \Rightarrow \cos t = \sqrt {{1 \over {{4 \over {19}}{{\left( {2x – {1 \over 2}} \right)}^2} + 1}}} = {1 \over {\sqrt {{4 \over {19}}{{\left( {2x – {1 \over 2}} \right)}^2} + 1} }} \cr & \Rightarrow F\left( x \right) = – \ln {1 \over {\sqrt {{4 \over {19}}{{\left( {2x – {1 \over 2}} \right)}^2} + 1} }} + C = \ln \sqrt {{4 \over {19}}{{\left( {2x – {1 \over 2}} \right)}^2} + 1} + C = {1 \over 2}\ln \left( {{4 \over {19}}{{\left( {2x – {1 \over 2}} \right)}^2} + 1} \right) + C \cr & = {1 \over 2}\ln \left[ {{4 \over {19}}\left( {{{\left( {2x – {1 \over 2}} \right)}^2} + {{19} \over 4}} \right)} \right] = {1 \over 2}\ln {4 \over {19}} + {1 \over 2}\ln \left[ {{{\left( {2x – {1 \over 2}} \right)}^2} + {{19} \over 4}} \right] + C \cr & = {1 \over 2}\ln \left( {4{x^2} – 2x + 5} \right) + C’\,\,\,\,\,\left( {C’ = C + {1 \over 2}\ln {4 \over {19}} = const} \right) \cr} \)

Chọn A.