Evaluate the following integrals :
`(i) int(x^(2)-4)/((x^(2)+1)(x^(2)+2)(x^(2)+3))dx` `(ii) int(1)/((x+1)(x^(2)+1)^(2))dx`
`(iii) int(1)/(x^(3)(x^(2)+1)^(2))dx`

Explanation

`(i)` We have,
`(x^(2)-4)/((x^(2)+1)(x^(2)+2)(x^(2)+3))=(A)/(x^(2)+1)+(B)/(x^(2)+2)+(C)/(x^(2)+3)`
`implies(t-4)/((t+1)(t+2)(t+3))=(A)/(t+1)+(B)/(t+2)+(C)/(t+3)`, for `=x^(2)`
`impliesA=(-5)/(2)`, `B=+6`, `C=(-7)/(2)`
`implies(x^(2)-4)/((x^(2)+1)(x^(2)+2)(x^(2)+3))=(-5)/(2)*(1)/(x^(2)+1)+(6)/(x^(2)+2)-(7)/(2)*(1)/(x^(2)+3)`
`impliesint(x^(2)-4)/((x^(2)+1)(x^(2)+2)(x^(2)+3))dx=(-5)/(2)tan^(-1)x+(6)/(sqrt(2))tan^(-1)((x)/(sqrt(2)))-(7)/(2)*(1)/(sqrt(3))tan^(-1)((x)/(sqrt(3)))+C`
`=(-5)/(2)tan^(-1)x+3sqrt(2)tan^(-1)((x)/(sqrt(2)))-(7)/(2sqrt(3))tan^(-1)((x)/(sqrt(3)))+C`
`(ii)` We have
`(1)/((x+1)(x^(2)+1)^(2))=(A)/(x+1)+(Bx+C)/(x^(2)+1)+(px+q)/((x^(2)+1)^(2))`
`implies1=A(x^(2)+1)^(2)+(Bx+C)(x^(2)+1)(x+1)+(px+q)(x+1)`.
Equating the coefficients of `x^(4)`, `x^(3)`, `x^(2)`, `x` and constant term from both sides, we get `A+B=0`, `B+C=0`, `2A+B+C+p=0`, `B+C+p+q=0` and `A+C+q=0`
`impliesA=(1)/(4)`, `B=-(1)/(4)`, `C=(1)/(4)`, `p=-(1)/(2)`, `q=(1)/(2)`
Thus `int(1)/((x+1)(x^(2)+1)^(2))dx=(1)/(4)int(1)/(x+1)dx-(1)/(4)int(x-1)/(x^(2)+1)dx-(1)/(2)int(x-1)/((x^(2)+1)^(2))dx`
`=(1)/(4)ln|x+1|-(1)/(8)int(2x)/(x^(2)+1)dx+(1)/(4)int(1)/(x^(2)+1)dx-(1)/(4)int(2x)/((x^(2)+1)^(2))dx+(1)/(2)(dx)/((x^(2)+1)^(2))`
`=(1)/(4)ln|x+1|-(1)/(8)ln(x^(2)+1)+(1)/(4)tan^(-1)x+(1)/(4)*(1)/((x^(2)+1))+(1)/(2)int(1)/((x^(2)+1)^(2))dx`
Now `int(1)/((x^(2)+1)^(2))dx=int(sec^(2)theta)/(sec^(4)theta)d theta` [Putting `(x=tantheta)` so that `dx=sec^(2)theta d theta`]
`=int(1+cos2theta)/(2)d theta=(1)/(2)[theta+(sin2theta)/(2)]`
`=(1)/(2)(theta+sinthetacostheta)=(1)/(2)[tan^(-1)x+(x)/(x^(2)+1)]`
Hence `int(1)/((x+1)(x^(2)+1)^(2))dx=(1)/(4)ln|x+1|-(1)/(8)ln(x^(2)+1)+(1)/(2)tan^(-1)x+(x+1)/(4(x^(2)+))+C` , `C` is constant of integration.
`(iii)` We have,
`int(1)/(x^(3)(x^(2)+1)^(2))dx=(1)/(2)int(2x)/(x^(4)(x^(2)+1)^(2))dx`
Let us put `x^(2)+1=t`
`2xdx=dt`
`x^(2)=t-1`
`=(1)/(2)int(dt)/(t^(2)(t-1)^(2))`
Now `(1)/(t^(2)(t-1)^(2))=(p)/(t-1)+(q)/((t-1)^(2))+(r )/(t)+(s)/(t^(2))`
`implies1=p(t-1)t^(2)+qt^(2)+rt(t-1)^(2)+s(t-1)^(2)`
whence `p=-2`, `q=1`, `r=2`, `s=1`
Thus `intint(1)/(x^(3)(x^(2)+1)^(2))dx=(1)/(2)[int(-2)/(t-1)dt+int(1)/((t-1)^(2))dt+int(2)/(t)dt+int(1)/(t^(2))dt]`
`=-ln|t-1|+ln|t|-(1)/(2(t-1))-(1)/(2t)+C`
`=ln|(x^(2)+1)/(x^(2))|-(1)/(2x^(2))-(1)/(2(x^(2)+1))+C `