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Evaluate $\int \frac{\left ( x^{2}-1 \right )dx}{\left ( x^{4}+3x^{2}+1 \right )\tan^{-1}\left ( x+\frac{1}{x} \right )}$

Option 1)
$-ln\left | \tan^{-1}\left ( x-\frac{1}{x} \right ) \right |+c$

Option 2)
$ln\left | \tan^{-1}\left ( x-\frac{1}{x} \right ) \right |+c$

Option 3)
$ln\left | \tan^{-1}\left ( x+\frac{1}{x} \right ) \right |+c$

Option 4)
$-ln\left | \tan^{-1}\left ( x+\frac{1}{x} \right ) \right |+c$

Answers (1)

best_answer

As we learnt

Type of Integration by perfect square -

Integration in the form of

(i) $\int f(x+\frac{1}{x})(1-\frac{1}{x^{2}})dx$

(ii) $\int f(x-\frac{1}{x})(1+\frac{1}{x^{2}})dx$

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

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

(v) $\int \frac{(1\pm \frac{1}{x^{2}})dx}{x^{2}+\frac{1}{x^{2}}}$

(vi) $\int \frac{f(x)dx}{ax^{4}+2bx^{3}+cx^{2}+2bx+a}$

- wherein

(i) $\rightarrow$ put $(x+\frac{1}{x})=t$

(ii) $\rightarrow$ put $(x-\frac{1}{x})=t$

(iii) $\rightarrow$ put $(x^{2}+\frac{1}{x^{2}})=t$

(iv) $\rightarrow$ put $(x^{2}-\frac{1}{x^{2}})=t$

(v) $\rightarrow$ for $1+\frac{1}{x^{2}}$ put $x-\frac{1}{x}=t$

$\rightarrow$ for $1-\frac{1}{x^{2}}$ put $x+\frac{1}{x}=t$

(vi) $\rightarrow$ put $(x+\frac{1}{x})=t$ if $b\neq 0$

put $(x^{2}+\frac{1}{x^{2}})=t$ if $b= 0$

Integral can be written as

$\int {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)dx}}{{\left[ {{{\left( {x + \frac{1}{x}} \right)}^2} + 1} \right]{{\tan }^{ - 1}}\left( {x + \frac{1}{x}} \right)}}}$

$Let \: \left( {x + \frac{1}{x}} \right) = t$ .

Differentiating we get $\left( {1 - \frac{1}{{{x^2}}}} \right)dx=dt$

$Hence,I=\int {\frac{{dt}}{{\left( {{t^2} + 1} \right){{\tan }^{ - 1}}t}}}$

Now make one more substitution tan^-1t = u. Then

$\frac{{dt}}{{{t^2} + 1}} = du\: \: {\rm{ and \: \: }}{\rm{I = }}\int {\frac{{{\rm{du}}}}{{\rm{u}}} = \ln |u| + c}$

$I = \ln \left| {{{\tan }^{ - 1}}t} \right| + c = \ln \left| {{{\tan }^{ - 1}}\left( {x - \frac{1}{x}} \right)} \right| + c$

Option 1)

$-ln\left | \tan^{-1}\left ( x-\frac{1}{x} \right ) \right |+c$

Option 2)

$ln\left | \tan^{-1}\left ( x-\frac{1}{x} \right ) \right |+c$

Option 3)

$ln\left | \tan^{-1}\left ( x+\frac{1}{x} \right ) \right |+c$

Option 4)

$-ln\left | \tan^{-1}\left ( x+\frac{1}{x} \right ) \right |+c$

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gaurav

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