Question

$\huge\underline\mathfrak\pink{Question}$
IN THE ATTACHMENT.
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Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle \rm \int \frac{dx}{ {x}^{4} + {x}^{2} + 1}$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle \rm \int \frac{dx}{ {x}^{4} + {x}^{2} + 1}$

can be rewritten as

$\rm \: = \: \: \dfrac{1}{2} \displaystyle \rm \int \frac{2}{ {x}^{4} + {x}^{2} + 1} \: dx$

$\rm \: = \: \: \dfrac{1}{2} \displaystyle \rm \int \frac{1 + 1}{ {x}^{4} + {x}^{2} + 1} \: dx$

$\rm \: = \: \: \dfrac{1}{2} \displaystyle \rm \int \frac{1 + 1 + {x}^{2} - {x}^{2} }{ {x}^{4} + {x}^{2} + 1} \: dx$

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

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

$\rm \: = \: \: \dfrac{1}{2} \: I_1 - \dfrac{1}{2} \: I_2$

So,

$\rm :\longmapsto\:\rm \: I = \: \: \dfrac{1}{2} \: I_1 - \dfrac{1}{2} \: I_2 - - - (1)$

where,

$\rm :\longmapsto\:I_1 = \displaystyle \rm \int \frac{({x}^{2} + 1) }{ {x}^{4} + {x}^{2} + 1} \: dx$

and

$\rm :\longmapsto\:I_2 = \displaystyle \rm \int \frac{({x}^{2} - 1) }{ {x}^{4} + {x}^{2} + 1} \: dx$

Now,

Consider,

$\rm :\longmapsto\:I_1 = \displaystyle \rm \int \frac{({x}^{2} + 1) }{ {x}^{4} + {x}^{2} + 1} \: dx$

Divide numerator and denominator by x^2, we get

$\rm \: = \: \: \displaystyle \rm \int \frac{1 + \dfrac{1}{ {x}^{2} } }{ {x}^{2} + \dfrac{1}{ {x}^{2} } + 1 } dx$

can be rewritten as

$\rm \: = \: \: \displaystyle \rm \int \frac{1 + \dfrac{1}{ {x}^{2} } }{ {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 + 2 + 1 } dx$

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

Now, we use method of Substitution,

$\red{\rm :\longmapsto\:Put \: x - \dfrac{1}{x} = y}$

$\red{\rm :\longmapsto\:\bigg(1 + \dfrac{1}{ {x}^{2} } \bigg)dx = dy}$

So, above integral can be rewritten as

$\rm \: = \: \: \displaystyle \rm \int \frac{dy}{ {y}^{2} + 3}$

$\rm \: = \: \: \displaystyle \rm \int \frac{dy}{ {y}^{2} + {( \sqrt{3} )}^{2} }$

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

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

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

So,

$\rm :\longmapsto\:I_1 = \: \: \dfrac{1}{ \sqrt{3} } {tan}^{ - 1} \dfrac{ {x}^{2} - 1}{ \sqrt{3}x } + c - - - (2)$

Now, Consider

$\rm :\longmapsto\:I_2= \displaystyle \rm \int \frac{({x}^{2} - 1) }{ {x}^{4} + {x}^{2} + 1} \: dx$

Divide numerator and denominator by x^2, we get

$\rm \: = \: \: \displaystyle \rm \int \frac{1 - \dfrac{1}{ {x}^{2} } }{ {x}^{2} + \dfrac{1}{ {x}^{2} } + 1 } dx$

can be rewritten as

$\rm \: = \: \: \displaystyle \rm \int \frac{1 - \dfrac{1}{ {x}^{2} } }{ {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 - 2 + 1 } dx$

$\rm \: = \: \: \displaystyle \rm \int \frac{1 - \dfrac{1}{ {x}^{2} } }{ {\bigg(x + \dfrac{1}{x} \bigg) }^{2} - 1} dx$

Now, we use method of Substitution, we get

$\red{\rm :\longmapsto\:Put \: x + \dfrac{1}{x} = y}$

$\red{\rm :\longmapsto\:\bigg(1 - \dfrac{1}{ {x}^{2} } \bigg)dx = dy}$

So, above integral can be rewritten as

$\rm \: = \: \: \displaystyle \rm \int \frac{dy}{ {y}^{2} - 1}$

$\rm \: = \: \: \dfrac{1}{2 \times 1}log \: \bigg |\dfrac{y - 1}{y + 1} \bigg| + d$

$\rm \: = \: \: \dfrac{1}{2}log \: \bigg |\dfrac{x + \dfrac{1}{x} - 1}{x + \dfrac{1}{x} + 1} \bigg| + d$

$\rm \: = \: \: \dfrac{1}{2}log \: \bigg |\dfrac{ {x}^{2} + 1 - x}{ {x}^{2} + 1 + x} \bigg| + d$

Hence,

$\rm :\longmapsto\:I_2= \: \: \dfrac{1}{2}log \: \bigg |\dfrac{ {x}^{2} + 1 - x}{ {x}^{2} + 1 + x} \bigg| + d - - - (3)$

Now, on substituting the values from equation (2) and equation (3) in equation (1), we get .

${ \rm \: I = \dfrac{1}{2\sqrt{3} } {tan}^{ - 1} \dfrac{ {x}^{2} - 1}{ \sqrt{3}x } + c - \dfrac{1}{4}log \: \bigg |\dfrac{ {x}^{2} + 1 - x}{ {x}^{2} + 1 + x} \bigg| + d}$

Hence,

${ \rm \: I = \dfrac{1}{2\sqrt{3} } {tan}^{ - 1} \dfrac{ {x}^{2} - 1}{ \sqrt{3}x } - \dfrac{1}{4}log \: \bigg |\dfrac{ {x}^{2} + 1 - x}{ {x}^{2} + 1 + x} \bigg| + e}$

where, c + d = e

Formula used :-

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

$\boxed{ \rm \: \displaystyle \rm \int \frac{dx}{ {x}^{2} - {a}^{2}} = \frac{1}{2a}log | \frac{x - a}{x + a} | + c}$

$\boxed{ \rm \: \dfrac{d}{dx}x = 1}$

$\boxed{ \rm \: \dfrac{d}{dx} \frac{1}{x} = - \frac{1}{ {x}^{2} } }$

$\boxed{ \rm \: {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2}}$

$\boxed{ \rm \: {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2}}$

Answer 2

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