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प्रश्न
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
विकल्प
`1/2 (x - 4) sqrt(x^2 - 8x + 7) + 9 log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C`
`1/2 (x + 4) sqrt(x^2 - 8x + 7) + 9 log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C`
`1/2 (x - 4) sqrt(x^2 - 8x + 7) + 3sqrt2 log abs (x - 4 + sqrt((x^2 - 8x + 7))) + C`
`1/2 (x - 4) sqrt(x^2 - 8x + 7) - 9/2 log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C`
उत्तर
`int sqrt(x^2 - 8x + 7) dx` is equal to `underline(1/2 (x - 4) sqrt(x^2 - 8x + 7) - 9/2 log abs (x - 4 + sqrt(x^2 - 8x + 7)) + C)`.
Explanation:
`int sqrt(x^2 - 8x + 7) dx`
`= int sqrt(x^2 - 8x + 16 + 7 - 16) dx`
`= int sqrt((x - 4)^2 - 9) dx`
`= int sqrt(x^2 - a^2) dx = x/2 sqrt (x^2 - a^2) - a^2/2 log abs (x + sqrt(x^2 - a^2)) + C`
On substituting x - 4 and a2 = 9 in place of x,
`therefore int sqrt((x - 4)^2 - 9) dx = ((x - 4))/2 sqrt(x^2 - 8x + 7) - 9/2 log abs ((x - 4) + sqrt (x^2 - 8x + 7)) + C`
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