Evaluate: `intsin^(-1)((2x+2)/(sqrt(4x^2+8x+13)))dx`

Answer

`(x+1)"tan"^(-1)((2x+2)/(3))-(3)/(4)"log"(4x^(2)+8x+13)+c`

Explanation

Let `I=int sin^(-1)((2x+2)/(sqrt(4x^(2)+8x+13)))dx`
`=sin^(-1)((2x+2)/(sqrt((2x+2)^(2)+9)))dx`
Put `2x + 2 = 3 tan theta rArr 2 dx = 3 sec^(2) theta d theta`
`therefore I = int sin^(-1)((3 tan theta)/(sqrt(9 tan ^(2)theta+9)))*(3)/(2)sec^(2)theta d theta`
`=int sin^(-1)((3 tan theta)/(3 sec theta))*(3)/(2)sec^(2)theta d theta`
`=int sin^(-1)((sin theta)/(cos theta*sec theta))*(3)/(2)sec^(2)theta d theta`
`=(3)/(2)int sin^(-1)(sin theta)*sec^(2)theta d theta`
`=(3)/(2)int theta* sec^(2)theta d theta=(3)/(2)[theta*tan theta-int1*tan theta d theta]`
`=(3)/(2)[theta tan theta - log sec theta]+c`
`=(3)/(2)[tan^(-1)((2x+2)/(3))*((2x+2)/(3))-logsqrt(1+((2x+2)/(3))^(2))]+c_(1)`
`=(x+1)tan^(-1)((2x+2)/(3))-(3)/(4)log[1+((2x+2)/(3))^(2)]+c_(1)`
`=(x+1)tan^(-1)((2x+2)/(3))-(3)/(4)log(4x^(2)+8x+13)+c["let"(3)/(2)log 3+c_(1)=c]`