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Answer 1

➔ Given :

$\\ \red\bigstar \sf{I=\bigg \lmoustache \: \: \dfrac{1-x^{2}}{x(1-2x)} dx}$

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➔To Find:

• $\\ \bf{ The \: value \: of \: I=\bigg\lmoustache \: \: \dfrac{1-x^{2}}{x(1-2x)}=?}$

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➔Solution :

$\\ \implies\sf{I= \bigg\lmoustache \: \: \dfrac{1-x^{2}}{x(1-2x)}}$

$\\ \implies\sf{I=\bigg\lmoustache \: \: \dfrac{1}{2}+\dfrac{1}{2}\bigg(\dfrac{2-x}{x(1-2x)}\bigg) }$

•Let $\sf{\dfrac{2-x}{x(1-2x)}=\dfrac{A}{x}+\dfrac{B}{(1-2x) }}$

$\\ \sf{\big(2-x \big)=A \big(1-2x \big)+Bx \: \: \: \: ....(1)}$

•Substituting x=0 and $\bf{\dfrac{1}{2}}$ in equation (1) , we obtain

⇒ A=2 and B=3

$\\ \sf{\therefore \: \: \dfrac{2-x}{x(1-2x)}=\dfrac{2}{x}+\dfrac{3}{(1-2x)}}$

•Substituting in equation (1) , we obtain :

$\\ \\ \implies \sf \: \frac{1 - {x}^{2} }{x(1 - 2x)} = \frac{1}{2} + \frac{1}{2} \bigg \{ \frac{2}{x} + \frac{3}{(1 - 2x)} \bigg \}$

$\\ \\ \implies \sf \: \bigg \lmoustache \frac{1 - {x}^{2} }{x(1 - 2x)}dx = \bigg \lmoustache \: \bigg \{ \frac{1}{2} + \frac{1}{2} \bigg( \frac{2}{x} + \frac{3}{1 - 2x} \bigg) \bigg \}dx \\ \\ \\ \\ \implies \sf \: \bigg\lmoustache \frac{1 - {x}^{2} }{x(1 - 2x)} dx = \frac{x}{2} + log |x| + \frac{3}{2( - 2)} log |1 - 2x| + c \\ \\ \\ \\ \implies \sf \: \bigg \lmoustache \frac{1 - {x}^{2} }{x(1 - 2x)} dx = \frac{x}{2} + log |x| - \frac{3}{4} log |1 - 2x| + c$

$\\ \red\bigstar \boxed{\sf{\bigg\lmoustache \dfrac{1-x^{2}}{x(1-2x)}dx=\dfrac{x}{2}+log|x|-\dfrac{3}{4}log|1-2x|+c}}$

➔Integration formulas :

$\\ \bf{(1) \bigg\lmoustache \: 1 dx =x+C}$

$\\ \bf{(2) \bigg\lmoustache \: a \: dx=ax+C}$

$\\ \bf{(3) \bigg\lmoustache \: x^{n}dx=\dfrac{x^{n+1}}{n+1}+C}$

$\\ \bf{(4) \bigg\lmoustache \: sinx dx=-cosx+C}$

$\\ \bf{(5) \bigg\lmoustache \: cosx dx=sinx+C}$

$\\ \bf{(6) \bigg\lmoustache \: sec^{2}x dx=tanx+C}$

$\\ \bf{(7) \bigg\lmoustache \: cosec^{2}x=-cotx+C}$

$\\ \bf{(8) \bigg\lmoustache \: secx.tanx dx=secx+C}$

$\\ \bf{(9) \bigg\lmoustache \:cosecx(cotx) dx=-cosecx+C}$

$\\ \bf{(10) \bigg\lmoustache \: \dfrac{1}{x} dx=ln(x)+C}$

$\\ \bf{(11) \bigg\lmoustache \: e^{x}dx=e^{x}+C}$

$\\ \bf{(12) \bigg\lmoustache \: a^{x}dx=\dfrac{a^{x}}{lna}+C}$

$\\ \bf{(13) \bigg\lmoustache \: \dfrac{1}{\sqrt{1-x^{2}} }\: dx= sin^{-1}x+C}$

$\\ \bf{(14) \bigg\lmoustache \: \dfrac{1}{1+x^{2}}dx=tan^{-1}x}$