Question

integrate (x²+x-6)/[(x-2)(x+1)]​

Answer 1

Step-by-step explanation:

Considering ∫(x2 + x - 1)/(x2 + x - 6) dx By expressing the integral Read more on Sarthaks.com - https://www.sarthaks.com/647228/evaluate-the-integral-x-2-x-1-x-2-x-6-dx?show=647235#a647235

Answer 2

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

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{ {x}^{2} + x - 6}{(x - 2)(x + 1)} \: dx$

$\rm  \:  =  \: \:\displaystyle\int\sf \: \dfrac{ {x}^{2} + 3x - 2x - 6}{(x - 2)(x + 1)} \: dx$

$\rm  \:  =  \: \:\displaystyle\int\sf \: \dfrac{x(x + 3)- 2(x + 3)}{(x - 2)(x + 1)} \: dx$

$\rm  \:  =  \: \:\displaystyle\int\sf \: \dfrac{(x + 3)(x - 2)}{(x - 2)(x + 1)} \: dx$

$\rm  \:  =  \: \:\displaystyle\int\sf \: \dfrac{x + 3}{x + 1} \: dx$

$\rm  \:  =  \: \:\displaystyle\int\sf \: \dfrac{x + 1 + 2}{x + 1} \: dx$

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

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

$\rm  \:  =  \: \:x + 2 log(x + 1) + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{ {x}^{2} + x - 6}{(x - 2)(x + 1)} \: dx \: = \: x \: + 2 log(x + 1) + c$

Formula Used :-

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

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{1}{x} \: dx \: = \: log(x) + c$

Additional Information :-

$\rm :\longmapsto\:\displaystyle\int\sf \: sinx \: dx = - \: cosx + c$

$\rm :\longmapsto\:\displaystyle\int\sf \: cosx \: dx = \: sinx + c$

$\rm :\longmapsto\:\displaystyle\int\sf \: cosecx \: cotx\: dx = - \: cosecx + c$

$\rm :\longmapsto\:\displaystyle\int\sf \: secx \: tanx\: dx = \: secx + c$

$\rm :\longmapsto\:\displaystyle\int\sf \: {sec}^{2}x \: dx = tanx \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: {cosec}^{2}x \: dx = - \: cotx \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: tanx \: dx = log(secx) \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: cotx \: dx = log(sinx) \: + \: c$

$\rm :\longmapsto\:\displaystyle\int\sf \: secx \: dx = log(secx + tanx) \: + \: c$