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

Explanation

`(i)` We have,
`int(1)/((x+1)sqrt(2x-3))dx`
Let us put `2x-3=t^(2)` so that `dx=tdt` and `x=(t^(2)+3)/(2)impliesx+1=(t^(2)+5)/(2)`
`=int(tdt)/((t^(2)+5)/(2).t)=2int(1)/(t^(2)+5)dt=(2)/(sqrt(5))tan^(-1)((t)/(sqrt(5)))+C`
`==(2)/(sqrt(5))tan^(-1)((sqrt(2x-3))/(sqrt(5)))+C`
`(ii)` We have,
`(1)/((x+1)sqrt(x^(2)+x-1))dx`
Let us put `(1)/(x+1)=t` so that `x=(1)/(t)-1` and `dx=-(1)/(t^(2))dt`
`=int(t)/(sqrt((1)/(t^(2))-(2)/(t)+1+(1)/(t)-1-1))xx-(1)/(t^(2))dt`
`=-int(1)/(sqrt(1-t-t^(2)))dt=-int(1)/(sqrt((5)/(4)-(t+(1)/(2))^(2)))dt`
`=cos^(-1)((t+(1)/(2))/((sqrt(5))/(2)))+C`
`=cos^(-1)((2t+1)/(sqrt(5)))+C=cos^(-1)((x+3)/(sqrt(5)(x+1)))+C`
`(iii) int(1)/((x^(2)+3x+3)sqrt(x+1))dx`
Let us put `x+1=t^(2)` so that `dx=2tdt` and
`x^(2)+3x+3=(t^(2)-1)^(2)+3(t^(2)-1)+3`
`=t^(4)-2t^(2)+1+3t^(2)-3+3`
`=t^(4)+t^(2)+1`
`=int(1)/((t^(4)+t^(2)+1)*t)*2tdt`
`=int(2)/(t^(4)+t^(2)+1)dt=int((2)/(t^(2)))/(t^(2)+(1)/(t^(2))+1)dt`
`=int((1+(1)/(t^(2)))-(1-(1)/(t^(2))))/(t^(2)+(1)/(t^(2))+1)`
`=int(1+(1)/(t^(2)))/((t-(1)/(t))^(2)+3)dt-int(1-(1)/(t^(2)))/((t+(1)/(t))^(2)-1)dt`
`=int(d(t-(1)/(t)))/((t-(1)/(t))^(2)+3)-int(d(t+(1)/(t)))/((t+(1)/(t))^(2)-1)dt`
`=(1)/(sqrt(3))tan^(-1)((t-(1)/(t))/(sqrt(3)))-(1)/(2)ln|(t+(1)/(t)-1)/(t+(1)/(t)+1)|+C`
`=(1)/(sqrt(3))tan^(-1)((x)/(sqrt(3(x+1))))-(1)/(2)ln|((x+2)-sqrt(x+1))/((x+2)+sqrt(x+1))|+C`, where `C` is a constant of integration.
`(iv)` We have,
`int(1)/((x^(2)-3x+2)sqrt(x^(2)-2))dx=int(1)/((x-1)(x-2))*(1)/(sqrt(x^(2)-2))dx`
`=int((1)/(x-2)-(1)/(x-1))(1)/(sqrt(x^(2)-2))dx`
`=int(1)/((x-2)sqrt(x^(2)-2))dx-int(1)/((x-1)sqrt(x^(2)-2))dx`
`=l_(1)-l_(2)`,say
Now `l_(1)=int(1)/((x-2)sqrt(x^(2)-2))dx`
`=int(txxt)/(sqrt(2t^(2)+4t+1))xx-(1)/(t^(2))dt` Let us put `x-2=(1)/(t)`
`=(-1)/(sqrt(2))int(1)/(sqrt(t^(2)+2t+(1)/(2)))dt` `dx=-(1)/(t^(2))dt`
`=(-1)/(sqrt(2))int(1)/(sqrt((t+1)^(2)-(1)/(2)))dt` `x=(1)/(t)+2=(1+2t)/(t)`
`=-(1)/(sqrt(2))ln|(t+1)+sqrt(t^(2)+2t+(1)/(2))|+C_(1)` `x^(2)-2=(1+4t+4t^(2)-2t^(2))/(t^(2))`
`=-(1)/(sqrt(2))ln|(x-1)/(x-2)+(sqrt(x^(2)-2))/(sqrt(2)(x-2))|+C_(1)`,`C_(1)` is a constant of integration.
and `l_(2)=int(1)/((x-1)sqrt(x^(2)-2))dx`
Let us put `x-1=(1)/(u)`
`=int(uxxu)/(sqrt(1+2u-u^(2))xx-(1)/(u^(2))du` `dx=-(1)/(u^(2))du`
`=-int(1)/(sqrt(2-(u+1)^(2))du` `x^(2)-2=((1)/(u)+1)^(2)-2`
`=cos^(-1)((u-1)/(sqrt(2)))+C_(2)=cos^(-1)((x-2)/(sqrt(2)(x-1)))+C_(2)` `=(1+2u-u^(2))/(u^(2))`
Hence, `l=int(1)/((x^(2)-3x+2)sqrt((x^(2)-2)))dx=-(1)/(sqrt(2))ln|(x-1)/(x-2)+(sqrt(x^(2)-2)/(sqrt(2)(x-2))|-cos^(-1)((x-2)/(sqrt(2)(x-1)))+k` where `k` is a constant.