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

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(B) To solve the integral $\int \frac{2x+3}{(x-1)(x^2+1)} dx$,we use partial fractions: $\frac{2x+3}{(x-1)(x^2+1)} = \frac{B}{x-1} + \frac{Cx+D}{x^2+1

Vedclass · Accepted Answer

$-\frac{5}{4}$

Vedclass · Answer

(B) To solve the integral $\int \frac{2x+3}{(x-1)(x^2+1)} dx$,we use partial fractions: $\frac{2x+3}{(x-1)(x^2+1)} = \frac{B}{x-1} + \frac{Cx+D}{x^2+1}$ $2x+3 = B(x^2+1) + (Cx+D)(x-1)$ Setting $x=1$,we get $2(1)+3 = B(1^2+1) \implies 5 = 2B \implies B = \frac{5}{2}$. Comparing coefficients of $x^2$: $0 = B+C \implies C = -B = -\frac{5}{2}$. Comparing constants: $3 = B-D \implies D = B-3 = \frac{5}{2} - 3 = -\frac{1}{2}$. Thus,the integral becomes $\int \left( \frac{5/2}{x-1} + \frac{-5/2x - 1/2}{x^2+1} ight) dx$. $= \frac{5}{2} \int \frac{1}{x-1} dx - \frac{5}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$. $= \frac{5}{2} \log_e |x-1| - \frac{5}{4} \log_e (x^2+1) - \frac{1}{2} 	an^{-1} x + A$. $= \log_e |x-1|^{\frac{5}{2}} + \log_e (x^2+1)^{-\frac{5}{4}} - \frac{1}{2} 	an^{-1} x + A$. $= \log_e \{(x-1)^{\frac{5}{2}}(x^2+1)^{-\frac{5}{4}}\} - \frac{1}{2} 	an^{-1} x + A$. Comparing this with the given expression,we find $a = -\frac{5}{4}$.

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

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