Question

integration of 1+x+x^2/ x^2(x+1)​

Answer 1

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

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

can be rewritten as

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

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

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

$\rm \: \: = \: \:\displaystyle \int \:{ {x}^{ - 2}} \: dx + log(1 + x) + c$

$\red{\bigg \{ \because \:\displaystyle \int \: \frac{1}{x}dx = log(x) + c\bigg \}}$

$\rm \: \: = \: \:\displaystyle \frac{ {x}^{ - 2 + 1} }{ - 2 + 1} + log(1 + x) + c$

$\rm \: \: = \: \:\displaystyle \frac{ {x}^{ - 1} }{ - 1} + log(1 + x) + c$

$\rm \: \: = \: \:\displaystyle \frac{ - 1 }{x} + log(1 + x) + c$

$\rm \: \: = \: - \:\displaystyle \frac{1 }{x} + log(1 + x) + c$

Additional Information :-

$\red{\rm :\longmapsto\:\displaystyle \int \: {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} \: + c}$

$\red{\rm :\longmapsto\:\displaystyle \int \: kdx \: = kx \: + c}$

$\red{\rm :\longmapsto\:\displaystyle \int \: sinx \: dx \: = - \: cosx \: + c}$

$\red{\rm :\longmapsto\:\displaystyle \int \: cosx \: dx \: = \: sinx \: + c}$

$\red{\rm :\longmapsto\:\displaystyle \int \: cosecx \: dx \: = - \: cosex \: cotx\: + c}$

$\red{\rm :\longmapsto\:\displaystyle \int \: secx \: dx \: = \: sex \: tanx\: + c}$