Question

Integrate the following function with respect to x

$\frac{ {x}^{4} + 1 }{ {x}^{2} + 1 }$
​

Answer 1

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

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

can be rewritten as

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

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

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

We know,

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

So, using this result, we get

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

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

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

We know,

$\boxed{\rm{  \:\displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

So, using this, we get

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

Hence,

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

$\rule{190pt}{2pt}$

Additional Information :-

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

Answer 2

• please check the attached file