If int {frac{{2x + 3}}{{(x - 1)({x^2} + 1)}}d | vedclass

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(D) Let $I = \int {\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}}dx}$.Using partial fractions,we write $\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}} = \frac{A_1}{x -

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$-5/4$

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(D) Let $I = \int {\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}}dx}$.Using partial fractions,we write $\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}} = \frac{A_1}{x - 1} + \frac{{Bx + C}}{{{x^2} + 1}}$.Solving for constants,we get $2x + 3 = A_1({x^2} + 1) + (Bx + C)(x - 1)$.For $x = 1$,$5 = 2A_1 \implies A_1 = \frac{5}{2}$.Comparing coefficients of $x^2$,$0 = A_1 + B \implies B = -\frac{5}{2}$.Comparing constants,$3 = A_1 - C \implies C = A_1 - 3 = \frac{5}{2} - 3 = -\frac{1}{2}$.Thus,$I = \int {\frac{{5/2}}{{x - 1}}dx} + \int {\frac{{-5/2x - 1/2}}{{{x^2} + 1}}dx}$.$I = \frac{5}{2}\log |x - 1| - \frac{5}{4}\int {\frac{{2x}}{{1 + {x^2}}}dx} - \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}dx}$.$I = \frac{5}{2}\log |x - 1| - \frac{5}{4}\log (1 + {x^2}) - \frac{1}{2}{	an ^{ - 1}}x + A$.$I = \log \left\{ {{{(x - 1)}^{\frac{5}{2}}}{{({x^2} + 1)}^{ - 5/4}}} ight\} - \frac{1}{2}{	an ^{ - 1}}x + A$.Comparing with the given expression,$a = -\frac{5}{4}$.

If $\int {\frac{{2x + 3}}{{(x - 1)({x^2} + 1)}}dx = {{\log }_e}\left\{ {{{(x - 1)}^{\frac{5}{2}}}{{({x^2} + 1)}^a}} \right\}} - \frac{1}{2}{\tan ^{ - 1}}x + A$,where $A$ is any arbitrary constant,then the value of $a$ is

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