Question

integrate (x²-x+1)/(x²+x+1) with respect to x​

Answer 1

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

Given integral is

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

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{2} + x + 1 - 2x}{ {x}^{2} + x + 1} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{2} + x + 1}{ {x}^{2} + x + 1} \: dx - \displaystyle\int\rm \frac{2x}{ {x}^{2} + x + 1} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \: dx - \displaystyle\int\rm \frac{2x + 1 - 1}{ {x}^{2} + x + 1} \: dx$

$\rm \:  =  \:x - \displaystyle\int\rm \frac{2x + 1}{ {x}^{2} + x + 1} \: dx + \displaystyle\int\rm \frac{dx}{ {x}^{2} + x + 1}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: \displaystyle\int\rm \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c \: \: }}}$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \displaystyle\int\rm \frac{dx}{ {x}^{2} + x + \dfrac{1}{4} - \dfrac{1}{4} + 1}$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \displaystyle\int\rm \frac{dx}{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} + \dfrac{3}{4}}$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \displaystyle\int\rm \frac{dx}{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} + {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} }$

We know,

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

So, using this identity we get

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \dfrac{2}{ \sqrt{3} } {tan}^{ - 1} \dfrac{x + \dfrac{1}{2} }{\dfrac{ \sqrt{3} }{2} } + c$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \dfrac{2}{ \sqrt{3} } {tan}^{ - 1} \dfrac{\dfrac{2x + 1}{2} }{\dfrac{ \sqrt{3} }{2} } + c$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\dfrac{2x + 1}{ \sqrt{3} } + c$

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MORE TO KNOW

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a}log\bigg | \frac{x - a}{x + a} \bigg| + c \: }}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } \: = \: log\bigg |x + \sqrt{ {x}^{2} + {a}^{2} } \bigg| + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } \: = \: log\bigg |x + \sqrt{ {x}^{2} - {a}^{2} } \bigg| + c}}$

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

Answer 2

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

Given integral is

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

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{2} + x + 1 - 2x}{ {x}^{2} + x + 1} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{2} + x + 1}{ {x}^{2} + x + 1} \: dx - \displaystyle\int\rm \frac{2x}{ {x}^{2} + x + 1} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \: dx - \displaystyle\int\rm \frac{2x + 1 - 1}{ {x}^{2} + x + 1} \: dx$

$\rm \:  =  \:x - \displaystyle\int\rm \frac{2x + 1}{ {x}^{2} + x + 1} \: dx + \displaystyle\int\rm \frac{dx}{ {x}^{2} + x + 1}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: \displaystyle\int\rm \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c \: \: }}}$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \displaystyle\int\rm \frac{dx}{ {x}^{2} + x + \dfrac{1}{4} - \dfrac{1}{4} + 1}$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \displaystyle\int\rm \frac{dx}{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} + \dfrac{3}{4}}$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \displaystyle\int\rm \frac{dx}{ {\bigg[x + \dfrac{1}{2} \bigg]}^{2} + {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} }$

We know,

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

So, using this identity we get

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \dfrac{2}{ \sqrt{3} } {tan}^{ - 1} \dfrac{x + \dfrac{1}{2} }{\dfrac{ \sqrt{3} }{2} } + c$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \dfrac{2}{ \sqrt{3} } {tan}^{ - 1} \dfrac{\dfrac{2x + 1}{2} }{\dfrac{ \sqrt{3} }{2} } + c$

$\rm \:  =  \: x - log | {x}^{2} + x + 1| + \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\dfrac{2x + 1}{ \sqrt{3} } + c$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a}log\bigg | \frac{x - a}{x + a} \bigg| + c \: }}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } \: = \: log\bigg |x + \sqrt{ {x}^{2} + {a}^{2} } \bigg| + c}}$

$\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } \: = \: log\bigg |x + \sqrt{ {x}^{2} - {a}^{2} } \bigg| + c}}$

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