Question

intrigrat the rational functions:. (x^2+1)(x^2+2)÷(x^2+3)(x^2+4)​

Answer 1

Answer:

this is positive or negative

write the heading

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int \sf \:\dfrac{( {x}^{2} + 1)( {x}^{2} + 2)}{( {x}^{2} + 3)( {x}^{2} + 4)}dx$

$\rm :\longmapsto\:Let \: {x}^{2} = y - - (1)$

So, given expression reduced to

$\rm :\longmapsto\:\sf \:\dfrac{( {x}^{2} + 1)( {x}^{2} + 2)}{( {x}^{2} + 3)( {x}^{2} + 4)} = \dfrac{(y + 1)(y + 2)}{(y + 3)(y + 4)}$

Now, using Partial Fraction, we have

$\rm :\longmapsto\:\dfrac{(y + 1)(y + 2)}{(y + 3)(y + 4)} =1 + \dfrac{a}{y + 3} + \dfrac{b}{y + 4} - - - (2)$

$\rm :\longmapsto\:(y + 1)(y + 2) = (y + 3)(y +4) + a(y + 4) + b(y + 3) - - (3)$

On substituting y = - 3, we get

$\rm :\longmapsto\:( - 3 + 1)( - 3 + 2) = ( - 3 + 3)( - 3+4) + a( - 3 + 4) + b( - 3 + 3)$

$\rm :\longmapsto\:( - 2)( - 1) = a(1)$

$\bf\implies \:a = 2$

On substituting y = - 4, we get

$\rm :\longmapsto\:( - 4 + 1)( - 4 + 2) = ( - 4 + 3)( - 4+4) + a( - 4 + 4) + b( - 4 + 3)$

$\rm :\longmapsto\:( - 3)( - 2) = b( - 1)$

$\bf\implies \:b = - \: 6$

Now substituting the values of a and b in equation (2), we get

$\rm :\longmapsto\:\dfrac{(y + 1)(y + 2)}{(y + 3)(y + 4)} =1 + \dfrac{2}{y + 3} - \dfrac{6}{y + 4}$

Now substituting the value of y, from equation (1), we get

$\rm :\longmapsto\:\dfrac{( {x}^{2} + 1)( {x}^{2} + 2)}{( {x}^{2} + 3)( {x}^{2} + 4)} =1 + \dfrac{2}{ {x}^{2} + 3} - \dfrac{6}{ {x}^{2} + 4}$

Thus,

On integrating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\displaystyle\int \sf \: \dfrac{( {x}^{2} + 1)( {x}^{2} + 2)}{( {x}^{2} + 3)( {x}^{2} + 4)}dx =\displaystyle\int \sf \: 1dx + \displaystyle\int \sf \: \dfrac{2}{ {x}^{2} + 3}dx-\displaystyle\int \sf \: \dfrac{6}{ {x}^{2} + 4}dx$

$\: \rm \: \: =x+2 \displaystyle\int \sf \: \dfrac{1}{ {x}^{2} + {( \sqrt{3} )}^{2} }dx-6\displaystyle\int \sf \:\dfrac{1}{ {x}^{2} + {2}^{2} }dx$

$\rm \: = \: \: x + 2 \times \dfrac{1}{ \sqrt{3} } {tan}^{ - 1}\dfrac{x}{ \sqrt{3} } - 6 \times\dfrac{1}{2} {tan}^{ - 1}\dfrac{x}{2} + c$

$\: \: \: \: \: \: \: \: \: \boxed{ \because \sf \: \displaystyle\int \sf \: \dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c}$

$\rm \: = \: \: x + \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\dfrac{x}{ \sqrt{3} } - 3{tan}^{ - 1}\dfrac{x}{2} + c$

Additional Information :-

$\boxed{ \sf \: \displaystyle\int \sf \: \dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c}$

$\boxed{ \sf \: \displaystyle\int \sf \: \dfrac{dx}{ {x}^{2} - {a}^{2} } = \dfrac{1}{2a} log\dfrac{x + a}{x - a} + c}$