Question

integrate the following function with respect to x

$\dfrac{1}{x + \sqrt{x} }$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{x + \sqrt{x} }$

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\: \sqrt{x} = y}$

$\purple{\rm :\longmapsto\:x = {y}^{2}}$

$\purple{\rm :\longmapsto\:dx = 2y \: dy}$

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

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

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

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

$\rm \:  =  \: 2log |y + 1| + c$

$\rm \:  =  \: 2log | \sqrt{x} + 1| + c$

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle\int\rm \frac{dx}{x + \sqrt{x} } = 2log | \sqrt{x} + 1| + c}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{x + \sqrt{x} }$

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\: \sqrt{x} = y}$

$\purple{\rm :\longmapsto\:x = {y}^{2}}$

$\purple{\rm :\longmapsto\:dx = 2y \: dy}$

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

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

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

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

$\rm \:  =  \: 2log |y + 1| + c$

$\rm \:  =  \: 2log | \sqrt{x} + 1| + c$

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle\int\rm \frac{dx}{x + \sqrt{x} } = 2log | \sqrt{x} + 1| + c}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

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