Integrate the functions`(5x)/((x+1)(x^2+9)`

Explanation

`int(5x)/((x+1)(x^(2)+9))dx`
`" Let " (5x)/((x+1)(x^(2)+9))=(A)/(x+1) +(Bx+C)/(x^(2)+9)`
`rArr " " 5x=A(x^(2)+9)+(Bx+C) (x+1)`
`x=- 1 " then " -5 =A(1 +9)+ +0 rArr A= -(1)/(2)`
`x=0 " then " 0= 9A+C rArr c=(9)/(2)`
Equating the coefficients of `x^(2)`
`A+B=0`
`rArr " " B=-A=(1)/(2)`
`:. int (5x)/((x+1)(x^(2)+9))dx -int (((-1)/(2)))/((x+1)) dx+ int ((1)/(2)x+(9)/(2))/(x^(2)+9)dx`
`=-(1)/(2) int (1)/(x+1) dx+(1)/(2) int ((x+9)/(x^(2)+9)) dx`
`=-(1)/(2) log |x+1|+(1)/(2) .(1)/(2) int (2x)/(x^(2)+9)dx`
`+(9)/(2) int(1)/(x^(2)+9)dx`
`=-(1)/(2) log |x+1|+(1)/(4) log |x^(2)+9|`
`+(9)/(2). (1)/(3) tan^(-1) ((x)/(3))+C`
` =-(1)/(2) log |x+1| +(1)/(4) log|x^(2)+9|`
`+(3)/(2)tan^(-1) .((x)/(3))+C`