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Integrate the function x+2x2+2x+3

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प्रश्न

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

योग
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उत्तर

Let `I = int (x + 2)/ sqrt (x^2 + 2x + 3)  dx`

`1/2 int (2x + 4)/ sqrt (x^2 + 2x + 3)  dx`

`1/2 int  (2x + 2 + 2)/sqrt (x^2 + 2x + 3)  dx`

`1/2 int (2x + 2)/ sqrt(x^2 + 2x + 3)  dx + dx/sqrt (x^2 + 2x + 3)`

Let I = I1 + I2             ....(i)

Where `I_1 = 1/2 int (2x + 2)/sqrt (x^2 + 2x + 3)  dx`

Let x2 + 2x + 3 = t

⇒ (2x + 2) dx = dt

∴ `I_1 = 1/2 int dt/sqrtt `

`= 1/2 int t^(-1/2) dt = 1/2 xx 2t^(1/2)`

`= 1/2 xx 2 sqrt (x^2 + 2x + 3) + C_1`

`= sqrt (x^2 + 2x + 3) + C_1`        .....(ii)

Also,

`I_2 = int dx/sqrt (x^2 + 2x + 3)`

`= int dx/sqrt (x^2 + 2x + 1 - 1 + 3)`

`= dx/ sqrt ((x + 1)^2 + sqrt( (2)^2))`

`log |(x + 1) + sqrt ((x + 1)^2 + 2)|`

`log |(x + 1) + sqrt (x^2 + 2x + 3)| + C_2`      ....(iii)

Hence from (i), (ii) and (iii), we get

`I = sqrt (x^2 + 2x + 3) + log |(x + 1) + sqrt (x^2 + 2x + 3)| + C`

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अध्याय 7: Integrals - Exercise 7.4 [पृष्ठ ३१६]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.4 | Q 21 | पृष्ठ ३१६

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