Question

please solve this question
thank you​

Answer 1

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

Given integral is

$\rm :\longmapsto\:I \: = \: \displaystyle\int \dfrac{ {x}^{5} - {2x}^{3} - 2x}{ {x}^{2} + 1} \: dx$

On adding and Subtracting 'x' in numerator, we get

$\rm :\longmapsto\:I \: = \: \displaystyle\int \dfrac{ {x}^{5} - {2x}^{3} - 2x + x - x}{ {x}^{2} + 1} \: dx$

can be re-arranged as

$\rm :\longmapsto\:I \: = \: \displaystyle\int \dfrac{ {x}^{5} - {2x}^{3} + x -3 x}{ {x}^{2} + 1} \: dx$

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

$\rm :\longmapsto\:I \: = \: \displaystyle\int \dfrac{x {( {x}^{2} - 1) }^{2} -3 x}{ {x}^{2} + 1} \: dx$

We know,

$\boxed{ \bf{ \: {(x - y)}^{2} = {(x + y)}^{2} - 4xy}}$

So, using this

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

$\rm :\longmapsto\:I = \displaystyle\int \dfrac{x {( {x}^{2} + 1)}^{2} - 4 {x}^{3} - 3x}{ {x}^{2} + 1}$

$\rm :\longmapsto\:I = \displaystyle\int \dfrac{x {( {x}^{2} + 1)}^{2} - 4 {x}^{3} - 4x + x}{ {x}^{2} + 1}$

$\rm :\longmapsto\:I = \displaystyle\int \dfrac{x {( {x}^{2} + 1)}^{2} - 4x({x}^{2} + 1) + x}{ {x}^{2} + 1}$

$\rm :\longmapsto\:I = \displaystyle\int x( {x}^{2} + 1)dx - \displaystyle\int 4xdx + \displaystyle\int \dfrac{x}{ {x}^{2} + 1 }dx$

$\rm :\longmapsto\:I = \displaystyle\int ({x}^{3} + x)dx - \displaystyle\int 4xdx +\dfrac{1}{2} \displaystyle\int \dfrac{2x}{ {x}^{2} + 1 }dx$

We know,

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

and

$\boxed{ \bf{ \displaystyle\int \frac{f'(x)}{f(x)} dx\: = logf(x) + c }}$

So, on applying these formula's we get,

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

$\rm :\longmapsto\:I = \dfrac{ {x}^{4} }{4} - \dfrac{ {3x}^{2} }{2}+ \dfrac{1}{2}log | {x}^{2} + 1 | + c$

Additional Information :-

$\boxed{ \bf{ \displaystyle\int kdx\: = kx + c}}$

$\boxed{ \bf{ \displaystyle\int {e}^{x} dx\: = {e}^{x} + c}}$

$\boxed{ \bf{ \displaystyle\int \frac{1}{x} dx\: =log |x| + c}}$

$\boxed{ \bf{ \displaystyle\int cosx \: dx\: = sinx + c}}$

$\boxed{ \bf{ \displaystyle\int sinx \: dx\: = - cosx + c}}$

Aliter Method :-

The given integral is

$\rm :\longmapsto\:I \: = \: \displaystyle\int \dfrac{ {x}^{5} - {2x}^{3} - 2x}{ {x}^{2} + 1} \: dx$

Since, degree of numerator > degree of denominator.

Using Long division method,

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{3} - 3x\:\:}}}\\ {\underline{\sf{ {x}^{2} + 1}}}& {\sf{{x}^{5}  -  {2x}^{3} - 2x\:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  -  {x}^{5} - {x}^{3}   \: \: \: \: \:  \:  \:  \:  \:  \:  \:  \:  \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: -  {3x}^{3} - 2x \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \:  \:  \:  \:  \:  \: \: \: \: \: \: \: {3x}^{3}  + 3x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \:  x \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

So, using long division we get

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

can be rewritten as

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

$\rm :\longmapsto\:I = \dfrac{ {x}^{4} }{4} - \dfrac{ {3x}^{2} }{2}+ \dfrac{1}{2}log | {x}^{2} + 1 | + c$