Tính nguyên hàm (x+5)/9x^2-1)*dx

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Tính nguyên hàm \[I = \int {\frac{{x - 5}}{{{x^2} - 1}}{\rm{d}}x} \]

\[I = \frac{3}{2}\ln \left| {\frac{{x + 1}}{{x - 1}}} \right| + C.\]

\[I = \frac{3}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C.\]

\[I = \ln \left| {\frac{{{{\left( {x + 1} \right)}^3}}}{{{{\left( {x - 1} \right)}^2}}}} \right| + C.\]

\[I = \ln \left| {\frac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^3}}}} \right| + C.\]

Giải thích

Đáp án C

Ta có: \(\frac{{x - 5}}{{{x^2} - 1}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} = \frac{{\left( {A + B} \right)x + A - B}}{{{x^2} - 1}}\)

Đồng nhất 2 vế ta có: \(\left\{ \begin{array}{l}A + B = 1\\A - B = - 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = - 2\\B = 3\end{array} \right.\)

Suy ra \(I = \int {\left( {\frac{3}{{x + 1}} - \frac{2}{{x - 1}}} \right)d{\rm{x}}} = 3\ln \left| {x + 1} \right| - 2\ln \left| {x - 1} \right| + C = \ln \left| {\frac{{{{\left( {x + 1} \right)}^3}}}{{{{\left( {x - 1} \right)}^2}}}} \right| + C\).