Question

`int sqrt(x^2 - 8x + 7)  dx` is equal to:-

विकल्प

• `1/2(x - 4) sqrt(x^2 - 8x + 7) + 9 log|x - 4 + sqrt(x^2 - 8x + 7)| + c`

• `1/2(x - 4) sqrt(x^2 - 8x + 7) + 9 log|x + 4 + sqrt(x^2 - 8x + 7)| + c`

• `1/2(x - 4) sqrt(x^2 - 8x + 7) + 3sqrt(2) log|x - 4 + sqrt(x^2 - 8x + 7)| + c`

• `1/2(x - 4) sqrt(x^2 - 8x + 7) - 9/2 log|x - 4 + sqrt(x^2 - 8x + 7)| + c`

Answer 1

`1/2(x - 4) sqrt(x^2 - 8x + 7) - 9/2 log|x - 4 + sqrt(x^2 - 8x + 7)| + c`

Explanation:

`int sqrt(x^2 - 8x + 7)  dx`

= `int sqrt(x^2 - 8x + 16 + 7)  dx`

= `int sqrt((x  - 4^2 - 9))  dx`

We have, `int sqrt(x^2 - a^2)  dx`

= `x/2 sqrt(x^2 - a^2) - a^2/2 log|x + sqrt(x^2 - a^2)| + c`

Replace `x` by `x - 4, a^2 = 9, a = 3`

∴ `int sqrt((x - 4)^2 - 9x)  dx`

= `((x - 4))/2 sqrt(x^2 - 8x + 7) - 9/2 log |(x - 4) + sqrt(x^2 - 8x + 7)| + c`