int {frac{{{x^4}}}{{(x - 1)({x^2} + 1)}}dx} = | vedclass

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(A) We have $I = \int \frac{x^4}{(x-1)(x^2+1)} dx$. Using polynomial division or algebraic manipulation: $x^4 = (x^4-1) + 1 = (x^2-1)(x^2+1) + 1 = (x-

Vedclass · Accepted Answer

$\frac{{x(x + 2)}}{2} + \frac{{\log |x - 1|}}{2} - \frac{{\log ({x^2} + 1)}}{4} - \frac{{{{	an }^{ - 1}}x}}{2} + c$

Vedclass · Answer

(A) We have $I = \int \frac{x^4}{(x-1)(x^2+1)} dx$. Using polynomial division or algebraic manipulation: $x^4 = (x^4-1) + 1 = (x^2-1)(x^2+1) + 1 = (x-1)(x+1)(x^2+1) + 1$. So,$\frac{x^4}{(x-1)(x^2+1)} = (x+1) + \frac{1}{(x-1)(x^2+1)}$. Now,we use partial fractions for $\frac{1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$. $1 = A(x^2+1) + (Bx+C)(x-1)$. Setting $x=1$,we get $1 = 2A \implies A = 1/2$. Comparing coefficients: $x^2: A+B=0 \implies B = -1/2$. Constant: $A-C=1 \implies C = A-1 = 1/2 - 1 = -1/2$. Thus,$\frac{1}{(x-1)(x^2+1)} = \frac{1}{2(x-1)} - \frac{x+1}{2(x^2+1)} = \frac{1}{2(x-1)} - \frac{x}{2(x^2+1)} - \frac{1}{2(x^2+1)}$. Integrating: $\int (x+1) dx + \frac{1}{2} \int \frac{1}{x-1} dx - \frac{1}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$. $= \frac{x^2}{2} + x + \frac{1}{2} \log |x-1| - \frac{1}{4} \log (x^2+1) - \frac{1}{2} 	an^{-1} x + c$. $= \frac{x^2+2x}{2} + \frac{1}{2} \log |x-1| - \frac{1}{4} \log (x^2+1) - \frac{1}{2} 	an^{-1} x + c = \frac{x(x+2)}{2} + \frac{1}{2} \log |x-1| - \frac{1}{4} \log (x^2+1) - \frac{1}{2} 	an^{-1} x + c$.

$\int {\frac{{{x^4}}}{{(x - 1)({x^2} + 1)}}dx} = $

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