Integrate the functions`(x+2)/(sqrt(4x-x^2))`

Explanation

`" Let " I= int (x+2)/(sqrt(4x-x^(2)))dx`
`" Let " x+2 =A .(d)/(dx) (4x-x^(2))+B`
`rArr x+2 =A (4-2x)+B`
Comparing the coefficients of x
`1=-2A rArr A=(-1)/(2)`
comparing constant terms
`2=4A+B`
`rArr 2=-2 +B`
`rArr B=4`
`:. I = int (-(1)/(2) (4-2x)+4)/(sqrt(4x-x^(2)))dx " " underset(rArr (4-2x)dx=dt)underset(" Let "4x-x^(2)=t)(" for first integral ")`
` =-(1)/(2) int (4-2x)/(sqrt(4x-x^(2)))dx+4 int (1)/(sqrt(4x-x^(2)))`
`=-(1)/(2) int (1)/(sqrt(t))dt+4 int (1)/(sqrt(2^(2)-(x-2)^(2)))dx`
`=-(1)/(2) int t^(-1//2) dt+4 int(1)/(sqrt(2^(2)-(x-2)^(2)))dx`
`=-(1)/(2) (t^(1//2))/(1//2)+4 sin^(-1) ((x-2)/(2))+c`
`=-sqrt(4x-x^(2)) +4 sin^(-1) ((x-2)/(2))+c``" Let I "= int (x+2)/(sqrt(x^(2)+2x+3))dx`
`" Let " x+2 =A (d)/(dx)(x^(2)+2x+3) +B`
`rArr x+2 =A (2x+2) +B`
Comparing coefficients of x
`1=2A rArr A=(1)/(2).`
Comparing constant terms
`2=2A+B`
`rArr 2=1+B`
`rArr B=1`
`:. I= int ((1)/(2) (2x+2)+1)/(sqrt(x^(2)+2x+3))dx`
`=(1)/(2) int (2x+2)/(sqrt(x^(2)+2x+3))dx" "underset(rArr (2x+2)dx=dt)underset(" Let "x^(2)+2x+3=t)(" For first integral ")`
`+ int (1)/(sqrt(x^(2)+2x+3))dx`
`=(1)/(2) int (1)/(sqrt(t))dt + int (1)/(sqrt((x^(2)+2x+1)+2))dx`
`=(1)/(2) int t^(-1//2) dt+ int (1)/(sqrt((x+1)^(2)+(sqrt(2))^(2)))dx`
`=(1)/(2) .(t^(1//2))/(1//2) +log |x+1+sqrt((x+1)^(2)+(sqrt(2))^(2))|+c`
`=sqrt(x^(2)+2x+3)+log |x+1+sqrt(x^(2)+2x+3)|+c`