Question

Integration of 1/(x+2) (x+1)^1/2​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{(x + 2) \sqrt{x + 1} }$

Its a special case of integration called p - q form, where p and q both are linear equations. We use method of Substitution to evaluate such integral.

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

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

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:dx = 2ydy$

On substituting all these values in given integral, we get

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

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

$\rm \: \: = \: 2 \: {tan}^{ - 1}y \: + \: c$

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

Additional Information :-

Integral of the form,

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{p \sqrt{q} }$

4 - Cases arises which are as follow

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf p&\sf \: q&\sf \:substitute\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf linear&\sf quadratic&\sf \: Put \: q = y\\\\\sf quadratic&\sf linear&\sf \: Put \: q = y\\\\\sf linear &\sf quadratic&\sf \: Put \: p = \dfrac{1}{y} \\\\\sf quadratic&\sf quadratic&\sf \: x = \dfrac{1}{y} \\\frac{\qquad}{}&\frac{\qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

More results :-

$\red{ \boxed{ \sf{ \: \displaystyle\int\tt \dfrac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c }}}$

$\red{ \boxed{ \sf{ \: \displaystyle\int\tt \dfrac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a} \: log \: \frac{x - a}{x + a} + c }}}$

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

$\red{ \boxed{ \sf{\displaystyle\int\tt \: \dfrac{dx}{ \sqrt{ {x}^{2} - {a}^{2} }} = log | \: x + \sqrt{ {x}^{2} - {a}^{2} } \: | + c}}}$