Question

integrate x^2 + 3x+1/x^4-x^2+1​

Answer 1

Answer:

${ {x}^{2} + 3x + \frac{1}{ {x}^{4} } } - {x }^{2} + 1$

Answer 2

Step-by-step explanation:

$\sf \int \bigg( \frac{( {x}^{2} + 3x + 1)}{ ({x}^{4} - {x}^{2} + 1) }\bigg)$

$\sf \int \frac{( {x}^{2} + 1)}{ ({x}^{4} - {x}^{2} + 1) }dx + 3 \int \frac{x}{ {x}^{4} - {x}^{2} + 1} dx$

$\sf \: let \: \: I = I_{1} + 3I_{2}$

$\sf \: I_{1} = \int \frac{ {x}^{2} + 1 }{ {x}^{4} - {x}^{2} + 1 } dx$

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

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

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

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

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

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

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

$\sf \: Let \: (x - \frac{1}{x} ) = t$

$\sf \: then \: (1 + \frac{1}{ {x}^{2} })dx = dt$

$= \sf \: \int \frac{ dt}{ ( {t}^{2} + 1) }$

$\sf \: = \tan^{ - 1} (t)$

$\sf \: l_{1} = \tan^{ - 1} \bigg(x - \frac{1}{x} \bigg )$

$\sf \: Now \: \: calculation \: \: of \: \: l_{2}$

$\sf l_{2} = \int \frac{x}{ {x}^{4} - {x}^{2} + 1 } dx$

$\sf \: {x}^{2} = p$

$\sf \: 2xdx = dp$

$\sf \: xdx = \frac{dp}{2}$

$\sf l_{2} = \frac{1}{2} \int \frac{dp}{ {p}^{2} - p + 1 }$

$\sf l_{2} = \frac{1}{2} \int \frac{dp}{ {p}^{2} - 2. p .( \frac{1}{2}) + \frac{1}{ {2}^{2} } - \frac{1}{ {2}^{2} } + 1 }$

$\sf l_{2} = \frac{1}{2} \int \frac{dp}{ {(p - \frac{1}{2} )}^{2} +( \frac{ \sqrt{3} }{2}) ^{2} }$

$\sf l_{2} = \frac{1}{2} . \frac{2}{ \sqrt{3} } \tan^{ - 1} \bigg( \frac{p - \frac{1}{2} }{ \frac{2}{ \sqrt{3} } } \bigg)$

$\sf l_{2} = \frac{1}{ \sqrt{3} } . \tan^{ - 1} \bigg( \frac{ \sqrt{3} (2p - 1)}{4} \bigg)$

$\sf \: Replace \: \: p \: \: to \: \: {x}^{2}$

$\sf l_{2} = \frac{1}{ \sqrt{3} } . \tan^{ - 1} \bigg( \frac{ \sqrt{3} (2 {x}^{2} - 1)}{4} \bigg)$

RequiredAns

$\sf \: let \: \: I = I_{1} + 3I_{2}$

$\sf l = \sf \: \tan^{ - 1} \bigg(x - \frac{1}{x} \bigg ) + 3.\frac{1}{ \sqrt{3} } . \tan^{ - 1} \bigg( \frac{ \sqrt{3} (2 {x}^{2} - 1)}{4} \bigg)$

$\sf l = \sf \: \tan^{ - 1} \bigg(x - \frac{1}{x} \bigg ) + \sqrt{3} . \tan^{ - 1} \bigg( \frac{ \sqrt{3} (2 {x}^{2} - 1)}{4} \bigg) + c$