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Evaluate the integral by using substitution. ∫02dxx+4-x2

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प्रश्न

Evaluate the integral by using substitution.

`int_0^2 dx/(x + 4 - x^2)`

योग
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उत्तर

Let `I = int_0^2  dx/(x + 4 - x^2)`

`= int_0^2 dx/(4 - (x^2 - x))`

`= int_0^2 dx/(4 + 1/4 - (x - 1/2)^2)`

`= int_0^2 dx/((sqrt17/2)^2 - (x - 1/2)^2)`

`= 1/(2 xx sqrt17/2) [log  (sqrt17/2 + (x - 1/2))/(sqrt17/2 - (x - 1/2)}]_0^2`

`= 1/sqrt17 [log  (sqrt17 + 2x - 1)/(sqrt17 - 2x  + 1)]_0^2`

`= 1/sqrt17 [log  (sqrt17 + 3)/(sqrt17 - 3) - log  (sqrt17 - 1)/(sqrt17 + 1)]`

`= 1/sqrt17  log [(sqrt17 + 3)/(sqrt17 - 3) xx (sqrt17 + 1)/(sqrt17 - 1)]`

`= 1/sqrt17 log [(17 +3 + 3sqrt17 + sqrt17)/(17 + 3 - 3sqrt17 - sqrt17)]`

`= 1/sqrt17  log ((20 + 4sqrt17)/(20 - 4sqrt17))`

`= 1/sqrt17  log ((5 + sqrt17)/(5 - sqrt17))`

`= 1/sqrt17  log ((5 + sqrt17)/(5 - sqrt17) xx (5 + sqrt17)/(5 + sqrt17))`

`= 1/sqrt17  log [(25 + 17 + 10sqrt17)/(25 - 17)]`

`= 1/sqrt17  log  [(41 + 10 sqrt17)/8]`

`= 1/sqrt17  log [(21 + 5 sqrt17)/4]`

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अध्याय 7: Integrals - Exercise 7.10 [पृष्ठ ३४०]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.10 | Q 6 | पृष्ठ ३४०

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