Question

Integrate with respect to x

$\frac{ {x}^{3} + 1}{ {x}^{4} + 1 }$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {x}^{3} + x}{ {x}^{4} + 1} \: dx$

Let assume that

$\rm :\longmapsto\:I = \displaystyle\int\rm \frac{ {x}^{3} + x}{ {x}^{4} + 1} \: dx$

can be rewritten as

$\rm \: I = \: \displaystyle\int\rm \frac{ {x}^{3} }{ {x}^{4} + 1}dx \: + \: \displaystyle\int\rm \frac{x}{ {x}^{4} + 1}dx$

Let further assume that,

$\rm :\longmapsto\:I = I_1 + I_2$

where,

$\boxed{ \tt{ \: I_1 \: = \: \displaystyle\int\rm \frac{ {x}^{3} }{ {x}^{4} + 1} \: dx \: }} \\ \\ \bf \:and \\ \\ \boxed{ \tt{ \:I_2 \: = \: \displaystyle\int\rm \frac{x}{ {x}^{4} + 1} \: dx \: }}$

Now, Consider

$\rm :\longmapsto\:I_1 = \displaystyle\int\rm \frac{ {x}^{3} }{ {x}^{4} + 1} \: dx$

To evaluate this integral, we use Method of Substitution

So, Substitute

$\red{\rm :\longmapsto\: {x}^{4} + 1 = y}$

$\red{\rm :\longmapsto\: {4x}^{3}dx = dy}$

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

So, on substituting these values, we get

$\rm :\longmapsto\:I_1 = \dfrac{1}{4}\displaystyle\int\rm \frac{dy}{y}$

$\rm :\longmapsto\:I_1 = \dfrac{1}{4}log |y| + c_1$

$\rm :\longmapsto\:\boxed{ \tt{ \: I_1 = \dfrac{1}{4}log | {x}^{4} + 1| + c_1 \: }}$

Now, Consider

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

can be rewritten as

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

To solve this integral, we use Method of Substitution

So, Substitute

$\red{\rm :\longmapsto\: {x}^{2} + 1 = z}$

$\red{\rm :\longmapsto\:2x \: dx = dz}$

$\red{\rm :\longmapsto\:x \: dx = \dfrac{dz}{2} }$

So, on substituting, all these values, we get

$\rm :\longmapsto\:I_2 = \dfrac{1}{2}\displaystyle\int\rm \frac{dz}{ {z}^{2} + 1}$

$\rm :\longmapsto\:I_2 = \dfrac{1}{2} {tan}^{ - 1}z + c_2$

$\rm :\longmapsto\:\boxed{ \tt{ \: I_2 = \dfrac{1}{2} {tan}^{ - 1} {x}^{2} + c_2 \: }}$

Hence,

$\rm :\longmapsto\:I = \dfrac{1}{4}log | {x}^{4} + 1| + c_1 + \dfrac{1}{2} {tan}^{ - 1} {x}^{2} + c_2$

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

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More to know :-

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

Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{I=\displaystyle\int\dfrac{x^3+1}{x^4+1}\,dx}$

$\tt{I=\displaystyle\int\dfrac{x^3}{x^4+1}\,dx+\int\dfrac{1}{x^4+1}\,dx}$

$\tt{\implies\,I=I_{1}+I_{2}}$

$\bold{\rm{\bullet\,\,SOLVING\,\,\,\,I_{1}}}$

$\tt{I_{1}=\displaystyle\int\dfrac{x^3}{1+x^4}\,dx}$

$\tt{\implies\,I_{1}=\dfrac{1}{4}\displaystyle\int\dfrac{4\,x^3}{1+x^4}\,dx}$

Substitute $\sf{1+x^4=u}$

$\sf{\implies\,4x^3\,dx=du}$

So,

$\tt{\implies\,I_{1}=\dfrac{1}{4}\displaystyle\int\dfrac{du}{u}}$

$\tt{\implies\,I_{1}=\dfrac{1}{4}\,\ln|u|+c_{1}}$

$\tt{\implies\,I_{1}=\dfrac{1}{4}\,\ln\left|1+x^4\right|+c_{1}}$

$\bold{\rm{\bullet\,\,SOLVING\,\,\,\,I_{2}}}$

$\tt{I_{2}=\displaystyle\int\dfrac{1}{x^4+1}\,dx}$

$\sf{Put\,\,x=\dfrac{1}{t}\,\,\,\,\,\,...(1)}$

$\sf{\implies\,dx=-\dfrac{1}{t^2}\,dt}$

$\tt{\implies\,I_{2}=-\displaystyle\int\dfrac{1}{\dfrac{1}{t^4}+1}\,\dfrac{dt}{t^2}}$

$\tt{\implies\,I_{2}=-\displaystyle\int\dfrac{t^4}{1+t^4}\,\dfrac{dt}{t^2}}$

$\tt{\implies\,I_{2}=-\displaystyle\int\dfrac{t^2}{1+t^4}\,dt}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2}\displaystyle\int\dfrac{2t^2}{1+t^4}\,dt}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2}\displaystyle\int\dfrac{t^2+1+t^2-1}{1+t^4}\,dt}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2}\displaystyle\int\dfrac{t^2+1}{1+t^4}\,dt-\dfrac{1}{2}\int\dfrac{t^2-1}{1+t^4}\,dt}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2}\displaystyle\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}}\,dt-\dfrac{1}{2}\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}}\,dt}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2}\displaystyle\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2}\,dt-\dfrac{1}{2}\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2}\,dt}$

$\sf{Put\,\,\,t-\dfrac{1}{t}=u\,\,\,\,\,\,\,and\,\,\,\,\,\,\,t+\dfrac{1}{t}=v}$

$\sf{\implies\,\left(1-\dfrac{1}{t^2}\right)\,dt=du\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\left(1+\dfrac{1}{t^2}\right)\,dt=dv}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2}\displaystyle\int\dfrac{du}{\left(u\right)^2+2}-\dfrac{1}{2}\int\dfrac{dv}{\left(v\right)^2-2}}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2}\displaystyle\int\dfrac{du}{\left(u\right)^2+\left(\sqrt{2}\right)^2}-\dfrac{1}{2}\int\dfrac{dv}{\left(v\right)^2-\left(\sqrt{2}\right)^2}}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2\sqrt{2}}tan^{-1}\left(\dfrac{u}{\sqrt{2}}\right)-\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{v-\sqrt{2}}{v+\sqrt{2}}\right|+c_{2}}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2\sqrt{2}}tan^{-1}\left(\dfrac{t-\dfrac{1}{t}}{\sqrt{2}}\right)-\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{t+\dfrac{1}{t}-\sqrt{2}}{t+\dfrac{1}{t}+\sqrt{2}}\right|+c_{2}}$

From (1), we get,

$\tt{\implies\,I_{2}=-\dfrac{1}{2\sqrt{2}}tan^{-1}\left(\dfrac{\dfrac{1}{x}-x}{\sqrt{2}}\right)-\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{\dfrac{1}{x}+x-\sqrt{2}}{\dfrac{1}{x}+x+\sqrt{2}}\right|+c_{2}}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2\sqrt{2}}tan^{-1}\left(\dfrac{1-x^2}{\sqrt{2}\,x}\right)-\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{1+x^2-\sqrt{2}\,x}{1+x^2+\sqrt{2}\,x}\right|+c_{2}}$

$\tt{\implies\,I_{2}=-\dfrac{1}{2\sqrt{2}}tan^{-1}\left(\dfrac{1-x^2}{\sqrt{2}\,x}\right)-\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{x^2-\sqrt{2}\,x+1}{x^2+\sqrt{2}\,x+1}\right|+c_{2}}$

$\tt{\implies\,I_{2}=\dfrac{1}{2\sqrt{2}}tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right)+\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{x^2+\sqrt{2}\,x+1}{x^2-\sqrt{2}\,x+1}\right|+c_{2}}$

Now,

$\tt{I=I_{1}+I_{2}}$

$\tt{\implies\,I=\dfrac{1}{4}\,\ln|1+x^4|+c_{1}\,+\dfrac{1}{2\sqrt{2}}\,tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right)+\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{x^2+\sqrt{2}\,x+1}{x^2-\sqrt{2}\,x+1}\right|+c_{2}}$$\sf{Put\,\,\,C=c_{1}+c_{2}}$

So,

$\tt{I=\dfrac{1}{4}\,\ln|1+x^4|+\dfrac{1}{2\sqrt{2}}\,tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right)+\dfrac{1}{4\sqrt{2}}\,\ln\left|\dfrac{x^2+\sqrt{2}\,x+1}{x^2-\sqrt{2}\,x+1}\right|+C}$