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Integrate the function 5x+3x2+4x+10 - Mathematics

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प्रश्न

Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`

योग

उत्तर

Let `I =int (5x +3)/ sqrt (x^2 + 4x + 10) dx`

Put 5x + 3

`= A [d/dx (x^2 + 4x + 10)] +B`

5x + 3 

= A (2x + 4) + B            ... (i)

Comparing the coefficient of x in (i), we get

5 = 2A

⇒ A = `5/2`

Comparing the constant terms in (i), we get

3 = 4A + B

⇒  B = -7

`I = int (5/2 (2x + 4) + (-7))/ sqrt (x^2 + 4x + 10)  dx`

`= 5/2 int (2x + 4)/sqrt (x^2 + 4x + 10)  dx - 7 int dx/ sqrt (x^2 + 4x + 10)`

`= 5/2 I_1 - 7I_2`

∴ `I = 5/2 I_1 - 7I_2`                       ....(ii)

Now, `I_1 = int (2x + 4)/ sqrt (x^2 + 4x + 10)  dx`

Put x2 + 4x + 10 = t

⇒  (2x + 4) dx = dt

∴ `I_1 = int dt/sqrtt`

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

`= 2 sqrt (x^2 + 4x + 10) + C_1`            ....(iii)

and `I_2 = int dx/ sqrt(x^2 + 4x + 10) `

`= int dx / sqrt((x + 2)^2 + (sqrt( 6))^2`

`= log |x + 2 + sqrt ((x + 2)^2 + (sqrt (6))^2)|`

`= log |x + 2 + sqrt (x^2 + 4x + 10)| + C_2`           ....(iv)

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

`I = 5 sqrt (x^2 + 4x + 10) - 7 log |x + 2 + sqrt (x^2 + 4x + 10)| + C`

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अध्याय 7: Integrals - Exercise 7.4 [पृष्ठ ३१६]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 7 Integrals
Exercise 7.4 | Q 23 | पृष्ठ ३१६

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