Math, asked by Anonymous, 1 month ago


\  \textless \ br /\  \textgreater \ {\red{ \boxed{ ∫ {x}^{3}. \ { \tan }^{ - 1} x \: dx}}}\}

Give answer with step-by-step by explanation

Expecting Answers From

Modereters
Maths ArryaBhatta
Other Best Users


Spam=Report

Explained Answer=Brainliest.​

Answers

Answered by mathdude500
7

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

Given integral is

\rm :\longmapsto\:\displaystyle\int\rm  {x}^{3} {tan}^{ - 1}x \: dx

To evaluate this integral, we use Method of Integration by Parts.

Integration by parts is

 \\ \purple{\boxed{\tt{ \displaystyle\int\rm uvdx = u\displaystyle\int\rm vdx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}u\displaystyle\int\rm vdx \bigg]dx}}} \\

Here,

\rm :\longmapsto\:u \:  =  \: {tan}^{ - 1}x

\rm :\longmapsto\:v \:  =  \:  {x}^{3}

So, on substituting the values, we get

\rm \:  =  \: {tan}^{ - 1}x\displaystyle\int\rm  {x}^{3}dx -  \displaystyle\int\rm \bigg[\dfrac{d}{dx}{tan}^{ - 1}x\displaystyle\int\rm  {x}^{3} dx \bigg]dx

\rm \:  =  \: {tan}^{ - 1}x \: \dfrac{ {x}^{4} }{4}  -  \displaystyle\int\rm \bigg[\dfrac{1}{ {x}^{2}  + 1} \times  \frac{ {x}^{4} }{4} \bigg]dx

\rm \:  = \: \dfrac{ {x}^{4}{tan}^{ - 1}x }{4}  -  \dfrac{1}{4} \displaystyle\int\rm \bigg[\dfrac{ {x}^{4} }{ {x}^{2}  + 1}  \bigg]dx

\rm \:  = \: \dfrac{ {x}^{4}{tan}^{ - 1}x }{4}  -  \dfrac{1}{4} \displaystyle\int\rm \bigg[\dfrac{ {x}^{4}  - 1 + 1}{ {x}^{2}  + 1}  \bigg]dx

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

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

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

\rm \:  = \: \dfrac{ {x}^{4}{tan}^{ - 1}x }{4}  -  \dfrac{1}{4} \displaystyle\int\rm \bigg[ {x}^{2} - 1 +  \dfrac{1}{ {x}^{2}  + 1}  \bigg]dx

\rm \:  = \: \dfrac{ {x}^{4}{tan}^{ - 1}x }{4}  -  \dfrac{1}{4}  \bigg[ \dfrac{ {x}^{3} }{3}  - x + {tan}^{ - 1}x  \bigg] + c

\rm \:  = \: \dfrac{1}{4}  \bigg[ {x}^{4}{tan}^{ - 1}x  - \dfrac{ {x}^{3} }{3} + x  - {tan}^{ - 1}x  \bigg] + c

Hence,

\boxed{\tt{ \displaystyle\int\rm  {x}^{3}{tan}^{ - 1}xdx =\dfrac{1}{4}  \bigg[ {x}^{4}{tan}^{ - 1}x  - \dfrac{ {x}^{3} }{3} + x  - {tan}^{ - 1}x  \bigg] + c}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Formula Used :-

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

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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}

Answered by as3801504
11

 \underline{ \boxed{\mathbb{\red{answer \: in \: attachement}}}}

{ \boxed{\mathbb{\pink{hope \: it \: help \: full \: for \: you}}}}

Attachments:
Similar questions