int {frac{{{x^3} - x - 2}}{{(1 - {x^2})}},dx} | vedclass

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Question

(D) We are given the integral $I = \int {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\,dx}$.First,simplify the integrand by splitting the numerator:$I = \int

Vedclass · Accepted Answer

$\log \left( {\frac{{x - 1}}{{x + 1}}} \right) - \frac{{{x^2}}}{2} + c$

Vedclass · Answer

(D) We are given the integral $I = \int {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\,dx}$.First,simplify the integrand by splitting the numerator:$I = \int {\frac{{x({x^2} - 1) - 2}}{{(1 - {x^2})}}\,dx}$Since $x^2 - 1 = -(1 - x^2)$,we can write:$I = \int {\frac{{-x(1 - {x^2}) - 2}}{{(1 - {x^2})}}\,dx}$$I = \int {\left( -x - \frac{2}{{1 - {x^2}}} \right)\,dx}$$I = -\int {x\,dx} - 2\int {\frac{1}{{1 - {x^2}}}\,dx}$Using the standard integral formula $\int {\frac{1}{{{a^2} - {x^2}}}\,dx} = \frac{1}{{2a}}\log \left| {\frac{{a + x}}{{a - x}}} \right| + c$,for $a=1$:$I = -\frac{{{x^2}}}{2} - 2 \left( \frac{1}{2} \log \left| {\frac{{1 + x}}{{1 - x}}} \right| \right) + c$$I = -\frac{{{x^2}}}{2} - \log \left| {\frac{{1 + x}}{{1 - x}}} \right| + c$Since $-\log \left| {\frac{{1 + x}}{{1 - x}}} \right| = \log \left| {\frac{{1 - x}}{{1 + x}}} \right| = \log \left| {\frac{{x - 1}}{{x + 1}}} \right|$,we get:$I = \log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{{{x^2}}}{2} + c$.Thus,the correct option is $D$.

$\int {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\,dx} = $

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