Integrate the functions `(x+2)/(sqrt(x^2+2x)+3)`

Explanation

`" 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`