मराठी

If dxabC∫dx(x+2)(x2+1)=alog|1+x2|+btan-1x+15log|x+2|+C, then ______.

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प्रश्न

If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then ______.

पर्याय

  • a = `(-1)/10`, b = `(-2)/5` 

  • a = `1/10`, b = `- 2/5`

  • a = `(-1)/10`, b = `2/5`

  • a = `1/10`, b = `2/5`

MCQ
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उत्तर

If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then a = `(-1)/10`, b = `2/5`.

Explanation:

Given that, `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`

Now, I = `int "dx"/((x + 2)(x^2 + 1))`

`1/((x + 2)(x^2 + 1)) = "A"/(x + 2) + ("B"x + "C")/(x^2 + 1)`

⇒ 1 = A(x2 + 1) + (Bx + C)(x + 2)

⇒ 1 = (A + B)x2 + (2B + C)x + A + 2C

Comapring coefficient, we get

A + B = 0

A + 2C = 1

2B + C = 0

Solving we get A = `1/5`

B = `- 1/5`

And C = `2/5`

∴ `int "dx"/((x + 2)(x^2 + 1))`

= `1/5 int 1/(x + 2) "d"x + int (- 1/5 + 2/5)/(x^2 + 1) "d"x`

= `1/5 int 1/(x + 2) "d"x - 1/10 int (2x)/(1 + x^2) "d"x + 1/5 int 2/(1 + x^2) "d"x`

= `1/5 log|x + 2| - 1/10 log|1 + x^2| + 2/5 tan^-1x + "C"`

= `"a" log |1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`  ....(Given)

∴ a = `(-1)/10`, b = `2/5`.

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पाठ 7: Integrals - Exercise [पृष्ठ १६८]

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