Question

$\displaystyle \sf \red{\int \frac{ {x}^{2} - 2}{ {(x}^{4} + 5 {x}^{2} + 4) \: arctan( \frac{ {x}^{2} + 2}{x}) } \: dx}$​

Answer 1

Solution:—

Given integral is:—

$\rm :\longmapsto\:\displaystyle \sf\int\frac{ {x}^{2} - 2 }{( {x}^{4} + 5 {x}^{2} + 4) \arctan\bigg( \dfrac{ {x}^{2} + 2 }{x} \bigg)}$

$\rm \: = \: \displaystyle \sf\int\frac{ {x}^{2} - 2 }{( {x}^{4} + 5 {x}^{2} + 4) {tan}^{ - 1} \bigg( \dfrac{ {x}^{2} + 2 }{x} \bigg)}$

$\rm \: = \: \displaystyle \sf\int\frac{ {x}^{2} - 2 }{( {x}^{4} + 5 {x}^{2} + 4) {tan}^{ - 1} \bigg( \dfrac{ {x}^{2} + 2 }{x} \bigg)}$

To evaluate this integral, we use method of Substitution:—

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

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

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

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

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

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

$\red{\rm :\longmapsto\:\dfrac{ {2x}^{2} - {x}^{2} - 2}{ {x}^{4} + {5x}^{2} + 4} = \dfrac{dy}{dx}}$

$\pink{\rm :\longmapsto\:\dfrac{ {x}^{2}- 2}{ {x}^{4} + {5x}^{2} + 4} = \dfrac{dy}{dx}}$

$\green{\rm :\longmapsto\:\dfrac{ {x}^{2}- 2}{ {x}^{4} + {5x}^{2} + 4}dx = dy}$

So, on substituting the values in above integral, we get :—

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

$\rm \: = \: log |y| + c$

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

Hence,

$\boxed{\tt{ \displaystyle \sf\int\frac{ {x}^{2} - 2 }{( {x}^{4} + 5 {x}^{2} + 4) {tan}^{ - 1} \bigg(\dfrac{{x}^{2} + 2}{x} \bigg)}dx = \: log\bigg | {tan}^{ - 1}\dfrac{ {x}^{2} + 2}{x} \bigg| + c}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Formula Used,

$\boxed{\tt{ \dfrac{d}{dx} {tan}^{ - 1}x = \frac{1}{1 + {x}^{2} }}}$

$\boxed{\tt{ \dfrac{d}{dx}k = 0}}$

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$

$\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} }}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information,

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