Question

Integrate the following function with respect to x

$\frac{ {x}^{2} - 1}{( {x}^{4} + 3 {x}^{2} + 1) {tan}^{ - 1} (x + \frac{1}{x} )}$
​

Answer 1

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

Given integral is

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

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{2} - 1}{( {x}^{4} + 3 {x}^{2} + 1) {tan}^{ - 1}\bigg(\dfrac{ {x}^{2} + 1}{x} \bigg) } \: dx$

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\: {tan}^{ - 1}\dfrac{ {x}^{2} + 1}{x} = y}$

So,

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {tan}^{ - 1}\dfrac{ {x}^{2} + 1}{x} = \dfrac{d}{dx}y}$

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

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

$\red{\rm :\longmapsto\:\dfrac{1}{ {x}^{2} + {x}^{4} + 1 + {2x}^{2} }\bigg[x(2x + 0) - ( {x}^{2} + 1)1 \bigg] \: = \dfrac{dy}{dx} }$

$\red{\rm :\longmapsto\:\dfrac{1}{{x}^{4} + 1 +{3x}^{2} }\bigg[ {2x}^{2} - {x}^{2} - 1 \bigg] \: = \dfrac{dy}{dx} }$

$\red{\rm :\longmapsto\:\dfrac{ {x}^{2} - 1}{{x}^{4} + 1 +{3x}^{2}}dx = dy }$

So, on comparing the values in given integral, we gat

$\rm \:  =  \: \displaystyle\int\rm \frac{dy}{y}$

$\rm \:  =  \: log |y| + c$

$\rm \:  =  \: log \bigg| {tan}^{ - 1}\bigg(\dfrac{ {x}^{2} + 1}{x} \bigg)\bigg| + c$

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More to know

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$