Question

Integrate the given function with respect to x

$\frac{ \sqrt{1 + {x}^{2} } }{x}$
​

Answer 1

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

Given integral is

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

To solve this integral, we use method of Substitution.

So, Substitute

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

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

$\red{\rm \: 2x \: dx = 2y \: dy \: }$

$\red{\rm \: x \: dx = y \: dy \: }$

$\red{\rm \: \sqrt{ {y}^{2} - 1} \: dx = y \: dy \: }$

$\red{\rm \: dx = \dfrac{ydy}{ \sqrt{ {y}^{2} - 1} } \: }$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle\int\rm \frac{y}{ \sqrt{ {y}^{2} - 1 } } \times \frac{y}{ \sqrt{ {y}^{2} - 1} } \: dy$

$\rm \: = \: \displaystyle\int\rm \frac{ {y}^{2} }{ {y}^{2} - 1} \: dy$

$\rm \: = \: \displaystyle\int\rm \frac{ {y}^{2} - 1 + 1}{ {y}^{2} - 1} \: dy$

$\rm \: = \: \displaystyle\int\rm \bigg(1 + \frac{ 1}{ {y}^{2} - 1}\bigg) \: dy$

$\rm \: = \: \displaystyle\int\rm dy + \displaystyle\int\rm \frac{dy}{ {y}^{2} - {1}^{2} }$

$\rm \: = \: y + \dfrac{1}{2 \times 1} log\bigg |\dfrac{y - 1}{y + 1} \bigg| + c$

$\rm \: = \: y + \dfrac{1}{2} log\bigg |\dfrac{y - 1}{y + 1} \bigg| + c$

$\rm \: = \: \sqrt{ {x}^{2} + 1} + \dfrac{1}{2} log\bigg |\dfrac{\sqrt{ {x}^{2} + 1} - 1}{\sqrt{ {x}^{2} + 1} + 1} \bigg| + c$

Hence,

$\\ \rm \: \displaystyle\int\rm \frac{\sqrt{ {x}^{2} + 1}}{x}dx = \: \sqrt{ {x}^{2} + 1} + \dfrac{1}{2} log\bigg |\dfrac{\sqrt{ {x}^{2} + 1} - 1}{\sqrt{ {x}^{2} + 1} + 1} \bigg| + c$

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

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2}} | + c}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

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

Answer 2

Answer:

$\large\underline{\sf{Solution-}} </p><p>Solution−</p><p></p><p> </p><p></p><p>Given integral is</p><p></p><p>\begin{gathered}\rm \: \displaystyle\int\rm \frac{ \sqrt{1 + {x}^{2} } }{x} \: dx \\ \end{gathered} </p><p>∫ </p><p>x</p><p>1+x </p><p>2</p><p> </p><p></p><p> </p><p></p><p> dx</p><p></p><p> </p><p></p><p>To solve this integral, we use method of Substitution.</p><p></p><p>So, Substitute</p><p></p><p>\begin{gathered} \red{\rm \: \sqrt{1 + {x}^{2} } = y} \\ \end{gathered} </p><p>1+x </p><p>2</p><p> </p><p></p><p> =y</p><p></p><p> </p><p></p><p>\begin{gathered} \red{\rm \: 1 + {x}^{2} = {y}^{2} \: } \\ \end{gathered} </p><p>1+x </p><p>2</p><p> =y </p><p>2</p><p> </p><p></p><p> </p><p></p><p>\begin{gathered} \red{\rm \: 2x \: dx = 2y \: dy \: } \\ \end{gathered} </p><p>2xdx=2ydy</p><p></p><p> </p><p></p><p>\begin{gathered} \red{\rm \: x \: dx = y \: dy \: } \\ \end{gathered} </p><p>xdx=ydy</p><p></p><p> </p><p></p><p>\begin{gathered} \red{\rm \: \sqrt{ {y}^{2} - 1} \: dx = y \: dy \: } \\ \end{gathered} </p><p>y </p><p>2</p><p> −1</p><p></p><p> dx=ydy</p><p></p><p> </p><p></p><p>\begin{gathered} \red{\rm \: dx = \dfrac{ydy}{ \sqrt{ {y}^{2} - 1} } \: } \\ \end{gathered} </p><p>dx= </p><p>y </p><p>2</p><p> −1</p><p></p><p> </p><p>ydy</p><p></p><p> </p><p></p><p> </p><p></p><p>So, on substituting these values in above integral, we get</p><p></p><p>\begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{y}{ \sqrt{ {y}^{2} - 1 } } \times \frac{y}{ \sqrt{ {y}^{2} - 1} } \: dy \\ \end{gathered} </p><p>=∫ </p><p>y </p><p>2</p><p> −1</p><p></p><p> </p><p>y</p><p></p><p> × </p><p>y </p><p>2</p><p> −1</p><p></p><p> </p><p>y</p><p></p><p> dy</p><p></p><p> </p><p></p><p>\begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{ {y}^{2} }{ {y}^{2} - 1} \: dy \\ \end{gathered} </p><p>=∫ </p><p>y </p><p>2</p><p> −1</p><p>y </p><p>2</p><p> </p><p></p><p> dy</p><p></p><p> </p><p></p><p>\begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{ {y}^{2} - 1 + 1}{ {y}^{2} - 1} \: dy \\ \end{gathered} </p><p>=∫ </p><p>y </p><p>2</p><p> −1</p><p>y </p><p>2</p><p> −1+1</p><p></p><p> dy</p><p></p><p> </p><p></p><p>\begin{gathered}\rm \: = \: \displaystyle\int\rm \bigg(1 + \frac{ 1}{ {y}^{2} - 1}\bigg) \: dy \\ \end{gathered} </p><p>=∫(1+ </p><p>y </p><p>2</p><p> −1</p><p>1</p><p></p><p> )dy</p><p></p><p> </p><p></p><p>\begin{gathered}\rm \: = \: \displaystyle\int\rm dy + \displaystyle\int\rm \frac{dy}{ {y}^{2} - {1}^{2} } \\ \end{gathered} </p><p>=∫dy+∫ </p><p>y </p><p>2</p><p> −1 </p><p>2</p><p> </p><p>dy</p><p></p><p> </p><p></p><p> </p><p></p><p>\rm \: = \: y + \dfrac{1}{2 \times 1} log\bigg |\dfrac{y - 1}{y + 1} \bigg| + c=y+ </p><p>2×1</p><p>1</p><p></p><p> log </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> </p><p>y+1</p><p>y−1</p><p></p><p> </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> +c</p><p></p><p>\rm \: = \: y + \dfrac{1}{2} log\bigg |\dfrac{y - 1}{y + 1} \bigg| + c=y+ </p><p>2</p><p>1</p><p></p><p> log </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> </p><p>y+1</p><p>y−1</p><p></p><p> </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> +c</p><p></p><p>\rm \: = \: \sqrt{ {x}^{2} + 1} + \dfrac{1}{2} log\bigg |\dfrac{\sqrt{ {x}^{2} + 1} - 1}{\sqrt{ {x}^{2} + 1} + 1} \bigg| + c= </p><p>x </p><p>2</p><p> +1</p><p></p><p> + </p><p>2</p><p>1</p><p></p><p> log </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> </p><p>x </p><p>2</p><p> +1</p><p></p><p> +1</p><p>x </p><p>2</p><p> +1</p><p></p><p> −1</p><p></p><p> </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> +c</p><p></p><p>Hence,</p><p></p><p>\begin{gathered} \\ \rm \: \displaystyle\int\rm \frac{\sqrt{ {x}^{2} + 1}}{x}dx = \: \sqrt{ {x}^{2} + 1} + \dfrac{1}{2} log\bigg |\dfrac{\sqrt{ {x}^{2} + 1} - 1}{\sqrt{ {x}^{2} + 1} + 1} \bigg| + c \\ \end{gathered} </p><p>∫ </p><p>x</p><p>x </p><p>2</p><p> +1</p><p></p><p> </p><p></p><p> dx= </p><p>x </p><p>2</p><p> +1</p><p></p><p> + </p><p>2</p><p>1</p><p></p><p> log </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> </p><p>x </p><p>2</p><p> +1</p><p></p><p> +1</p><p>x </p><p>2</p><p> +1</p><p></p><p> −1</p><p></p><p> </p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p>∣</p><p></p><p> +c</p><p></p><p> </p><p></p><p>\rule{190pt}{2pt}</p><p></p><p>Additional Information :-</p><p></p><p>\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2}} | + c}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered} </p><p>MoreFormulae</p><p>MoreFormulae</p><p></p><p> </p><p>★∫ </p><p>x </p><p>2</p><p> +a </p><p>2</p><p> </p><p>dx</p><p></p><p> = </p><p>a</p><p>1</p><p></p><p> tan </p><p>−1</p><p> </p><p>a</p><p>x</p><p></p><p> +c</p><p>★∫ </p><p>x </p><p>2</p><p> −a </p><p>2</p><p> </p><p></p><p> </p><p>dx</p><p></p><p> =log∣x+ </p><p>x </p><p>2</p><p> −a </p><p>2</p><p> </p><p></p><p> ∣+c</p><p>★∫ </p><p>a </p><p>2</p><p> −x </p><p>2</p><p> </p><p></p><p> </p><p>dx</p><p></p><p> =sin </p><p>−1</p><p> </p><p>a</p><p>x</p><p></p><p> +c</p><p>★∫ </p><p>x </p><p>2</p><p> +a </p><p>2</p><p> </p><p></p><p> </p><p>dx</p><p></p><p> =log∣x+ </p><p>x </p><p>2</p><p> +a </p><p>2</p><p> </p><p></p><p> ∣+c</p><p></p><p> </p><p></p><p> </p><p></p><p> </p><p></p><p> </p><p></p><p> </p><p></p><p> </p><p></p><p>\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></p><p> </p><p></p><p> </p><p></p><p> </p><p></p><p> </p><p></p><p>$