Question

Evaluate:-
$\\ \red{ \boxed{\displaystyle \pmb{ \tt\int \: x^{\frac{13}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx}}}$
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༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

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​

Answer 1

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

Given integral is

$\rm \: \displaystyle{ \tt\int \: x^{\frac{13}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx}$

can be rewritten as

$\rm \: = \displaystyle{ \tt\int \: x^{5 + \frac{3}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx}$

can be further rewritten as

$\rm \: = \displaystyle{ \tt\int \: {x}^{5} x^{ \frac{3}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx}$

$\rm \: = \displaystyle{ \tt\int \: {( {x}^{ \frac{5}{2}}) }^{2} x^{ \frac{3}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx}$

Now, Substitute

$\rm \: \sqrt{1 + {x}^{ \frac{5}{2} } } = y$

$\rm \: 1 + {x}^{ \frac{5}{2} } = {y}^{2}$

$\rm \: {x}^{ \frac{5}{2} } = {y}^{2} - 1$

$\rm \: \frac{5}{2} {x}^{ \frac{3}{2} } dx = 2y \: dy$

$\rm \: {x}^{ \frac{3}{2} } dx = \frac{4y}{5} \: dy$

So, on substituting these values, we get

$\rm \: = \displaystyle{ \tt\int \: {( {y}^{2} - 1) }^{2} \times y \times \frac{4y}{5} dx}$

$\rm \: = \frac{4}{5} \displaystyle{ \tt\int \: {y}^{2} {( {y}^{2} - 1) }^{2} dx}$

$\rm \: = \frac{4}{5} \displaystyle{ \tt\int \: {y}^{2}( {y}^{4} + 1 - {2y}^{2}) dx}$

$\rm \: = \frac{4}{5} \displaystyle{ \tt\int \:( {y}^{6} + {y}^{2} - {2y}^{4}) dx}$

$\rm \: = \dfrac{4}{5}\bigg(\dfrac{ {y}^{7} }{7} + \dfrac{ {y}^{3} }{3} - \dfrac{ {2y}^{5} }{5} \bigg) + c$

$\rm \: = \dfrac{4}{5}\bigg(\dfrac{ {(\sqrt{1 + {x}^{ \frac{5}{2} } }) \: }^{7} }{7} + \dfrac{ {(\sqrt{1 + {x}^{ \frac{5}{2} } }) \: }^{3} }{3} - \dfrac{ {2(\sqrt{1 + {x}^{ \frac{5}{2} } }) \: }^{5} }{5} \bigg) + c$

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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

$\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =$

Given integral is

$\begin{gathered}\rm \: \displaystyle{ \tt\int \: x^{\frac{13}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx} \\ \end{gathered} </p><p>∫x </p><p>2</p><p>13</p><p></p><p> </p><p> (1+x </p><p>2</p><p>5</p><p></p><p> </p><p> ) </p><p>2</p><p>1</p><p> dx$

can be rewritten as

$\begin{gathered}\rm \: = \displaystyle{ \tt\int \: x^{5 + \frac{3}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx} \\ \end{gathered} </p><p>=∫x </p><p>5+ </p><p>2</p><p>3</p><p></p><p> </p><p> (1+x </p><p>2</p><p>5</p><p></p><p> </p><p> ) </p><p>2</p><p>1</p><p></p><p> </p><p> dx$can be further rewritten as

$\begin{gathered}\rm \: = \displaystyle{ \tt\int \: {x}^{5} x^{ \frac{3}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx} \\ \end{gathered} </p><p>=∫x </p><p>5</p><p> x </p><p>2</p><p>3</p><p></p><p> </p><p> (1+x </p><p>2</p><p>5</p><p></p><p> </p><p> ) </p><p>2</p><p>1</p><p></p><p> </p><p> dx$

$\begin{gathered}\rm \: = \displaystyle{ \tt\int \: {( {x}^{ \frac{5}{2}}) }^{2} x^{ \frac{3}{2}}\left(1+x^{\frac{5}{2}}\right)^{\frac{1}{2}}dx} \\ \end{gathered} </p><p>=∫(x </p><p>2</p><p>5</p><p></p><p> </p><p> ) </p><p>2</p><p> x </p><p>2</p><p>3</p><p></p><p> </p><p> (1+x </p><p>2</p><p>5</p><p></p><p> </p><p> ) </p><p>2</p><p>1</p><p></p><p> </p><p> dx$

Now, Substitute

$\begin{gathered}\rm \: \sqrt{1 + {x}^{ \frac{5}{2} } } = y \\ \end{gathered} </p><p>1+x </p><p>2</p><p>5</p><p></p><p> </p><p> </p><p></p><p> =y</p><p>$

$\begin{gathered}\rm \: 1 + {x}^{ \frac{5}{2} } = {y}^{2} \\ \end{gathered} </p><p>1+x </p><p>2</p><p>5</p><p></p><p> </p><p> =y </p><p>2</p><p>$

$\begin{gathered}\rm \: {x}^{ \frac{5}{2} } = {y}^{2} - 1 \\ \end{gathered} </p><p>x </p><p>2</p><p>5</p><p></p><p> </p><p> =y </p><p>2</p><p> −1$

$\begin{gathered}\rm \: \frac{5}{2} {x}^{ \frac{3}{2} } dx = 2y \: dy \\ \end{gathered} </p><p>2</p><p>5</p><p></p><p> x </p><p>2</p><p>3</p><p></p><p> </p><p> dx=2ydy</p><p>$

$\begin{gathered}\rm \: {x}^{ \frac{3}{2} } dx = \frac{4y}{5} \: dy \\ \end{gathered} </p><p>x </p><p>2</p><p>3</p><p></p><p> </p><p> dx= </p><p>5</p><p>4y</p><p></p><p> dy$

So, on substituting these values, we get

$\begin{gathered}\rm \: = \displaystyle{ \tt\int \: {( {y}^{2} - 1) }^{2} \times y \times \frac{4y}{5} dx} \\ \end{gathered} </p><p>=∫(y </p><p>2</p><p> −1) </p><p>2</p><p> ×y× </p><p>5</p><p>4y</p><p></p><p> dx</p><p>$

$\begin{gathered}\rm \: = \frac{4}{5} \displaystyle{ \tt\int \: {y}^{2} {( {y}^{2} - 1) }^{2} dx} \\ \end{gathered} </p><p>= </p><p>5</p><p>4</p><p></p><p> ∫y </p><p>2</p><p> (y </p><p>2</p><p> −1) </p><p>2</p><p> dx$

$\begin{gathered}\rm \: = \frac{4}{5} \displaystyle{ \tt\int \: {y}^{2}( {y}^{4} + 1 - {2y}^{2}) dx} \\ \end{gathered} </p><p>= </p><p>5</p><p>4</p><p></p><p> ∫y </p><p>2</p><p> (y </p><p>4</p><p> +1−2y </p><p>2</p><p> )dx$

$\begin{gathered}\rm \: = \frac{4}{5} \displaystyle{ \tt\int \:( {y}^{6} + {y}^{2} - {2y}^{4}) dx} \\ \end{gathered} </p><p>= </p><p>5</p><p>4</p><p></p><p> ∫(y </p><p>6</p><p> +y </p><p>2</p><p> −2y </p><p>4</p><p> )dx$

$\begin{gathered}\rm \: = \dfrac{4}{5}\bigg(\dfrac{ {y}^{7} }{7} + \dfrac{ {y}^{3} }{3} - \dfrac{ {2y}^{5} }{5} \bigg) + c \\ \end{gathered} </p><p>= </p><p>5</p><p>4</p><p></p><p> ( </p><p>7</p><p>y </p><p>7</p><p> </p><p></p><p> + </p><p>3</p><p>y </p><p>3</p><p> </p><p></p><p> − </p><p>5</p><p>2y </p><p>5</p><p> </p><p></p><p> )+c</p><p>$

Additional Information :-

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