Question

Integration of 5x/(x²-2x+2) (x+1) dx

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \sf{ \int \dfrac{5x}{ \left( {x}^{2} - 2x + 2 \right) \left(x + 1 \right)} \: dx}$

Decomposing into partial fraction,

$\rm{ \dfrac{5x}{ \left( {x}^{2} - 2x + 2 \right)\left( x + 1 \right)} = \dfrac{Ax + B}{ {x}^{2} - 2x + 2 } + \dfrac{C}{x + 1} }$

$\rm{ \implies\dfrac{5x}{ \left( {x}^{2} - 2x + 2 \right)\left( x + 1 \right)} = \dfrac{ \left(Ax + B \right) \left(x + 1 \right) + C \left({x}^{2} - 2x + 2 \right)}{ \left({x}^{2} - 2x + 2 \right) \left( x + 1\right)} }$

$\rm{ \implies5x= Ax ^{2} + Bx + Ax + B+ C {x}^{2} - 2Cx + 2C }$

$\rm{ \implies5x= (A + C)x ^{2} + (A + B - 2C)x + (B + 2C ) }$

On comparing the coefficients,

$\rm{ \implies A + C = 0 \: \: \: \: , \: \: \: \: A + B - 2C = 5 \: \: \: \: , \: \: \: \: B + 2C = 0 }$

On solving these, we get,

$\rm{ \implies A = 1 \: \: \: \: , \: \: \: \: B = 2 \: \: \: \: , \: \: \: \: C = - 1 }$

So,

$\displaystyle \sf{ \int \dfrac{x + 2}{ {x}^{2} - 2x + 2} \: dx - \int \dfrac{dx}{x + 1}}$

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

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

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

$\displaystyle = \sf{ \dfrac{1}{2} \ln| {x}^{2} - 2x + 2| +3 \int\dfrac{dx }{ {(x - 1)}^{2} + 1} - \int \dfrac{dx}{x + 1}}$

$\displaystyle = \sf{ \dfrac{1}{2} \ln| {x}^{2} - 2x + 2| +3 \: tan^{ - 1} (x - 1)- \ln | x + 1 | + c}$

$\displaystyle = \sf{ \ln \left| \sqrt{{x}^{2} - 2x + 2 } \right| +3 \: tan^{ - 1} (x - 1)- \ln | x + 1 | + c}$

$\displaystyle = \sf{ \ln \left| \dfrac{ \sqrt{{x}^{2} - 2x + 2 }}{x + 1} \right| +3 \: tan^{ - 1} (x - 1) + c}$