Question

Integrate the following function with respect to x

$\frac{1}{x + 1 + \sqrt{x + 1} }$

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EVALUATE

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Answer 1

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

Given integral is

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

can be rewritten as

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

To evaluate this integral, we use method of Substitution.

So, we substitute

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

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

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

So,

On substituting these values in given integral, we get

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

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

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

We know

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\int\sf \frac{dx}{x} \: = \: logx \: + \: c \: }}}$

So, using this, we get

$\rm \: = \: 2 \: log |y + 1| + c$

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

Hence,

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

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Additional Information :-

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