Mathematics

# $\int {\dfrac{{5{x}}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}} \,\,dx$

##### SOLUTION
$\int\dfrac{5x}{(x+1)(x^2-4)}$

$\dfrac{5x}{(x+1)(x^2-4)}=\dfrac{5x}{(x+1)(x-2)(x+2)}$

$\dfrac{5x}{(x+1)(x^2-4)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x+2)}$

$\dfrac{5x}{(x+1)(x^2-4)}=\dfrac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)}$

By cancelling denominator
$5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)$
Putting $x=-1$, in ( 1 )
$5\times (-1)=A(-1-2)(-1+2)+B(-1+1)(-1+2)+C(-1+1)(-1-2)$
$-5=A(-3)(1)$
$\Rightarrow$  $A=\dfrac{5}{3}$
Similarly, putting $x=2,$ in ( 1 )
$5(2)=A(2-2)(2+2)+B(2+1)(2+2)+c(2+1)(2-2)$
$10=A\times 0+B(3)(4)+C\times 0$
$10=12B$
$\Rightarrow$  $B=\dfrac{10}{12}=\dfrac{5}{6}$
Similarly putting $x=-2,$ in ( 1 )
$5(-2)=A(-2-2)(-2+2)+B(-2+1)(-2+2)+C(-2+1)(-2-2)$
$-10=A\times 0+\times 0+C(-1)(-4)$
$-10=4C$
$C=\dfrac{-10}{4}=\dfrac{-5}{2}$
Therefore
$\int\dfrac{5x}{(x+1)(x^2-4)}=\int\left(\dfrac{A}{x+1}+\dfrac{B}{x-2}+\dfrac{C}{x+2}\right)dx$

$=\dfrac{5}{3}\int\dfrac{dx}{x+1}dx+\dfrac{5}{6}\int\dfrac{dx}{x-2}dx-\dfrac{5}{2}\int\dfrac{dx}{x+2}$

$=\dfrac{5}{3}log|x+1|-\dfrac{5}{2}\log|x+2|+\log|x-2|+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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