Question

Integrate the following w.r.t x
$\bf \dfrac{x}{\sqrt{1+x^2}}$

Answer 1

Integrate the following w.r.t x

x/√1+x²

Answer:—

→ ∫x/√(1+x²)dx

=½∫(2x)(1+x²)^(-½)dx

=½[(1+x²)^(½)/½]+C

=√(1+x²)+C

$\:$

Answer 2

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

Given integral is

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

To evaluate, this integral, we use method of Substitution.

Let we substitute,

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

On squaring both sides, we get

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

On differentiating, we get

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

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

So, on substituting these values, the given integral reduces to

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

$\rm = \: \displaystyle \int \: dy$

$\rm = \: y \: + \: c$

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

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle \int \frac{x}{ \sqrt{1 + {x}^{2} } } \: dx = \sqrt{1 + {x}^{2} } + c\: }}$

Additional Information :-

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