हिंदी

Find ∫x+2x2-4x-5dx.

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प्रश्न

Find `int (x + 2)/sqrt(x^2 - 4x - 5) dx`.

योग
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उत्तर

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

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

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

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

Let x2 – 4x – 5 = u

(2x – 4) = `(du)/dx`

(x – 2) dx = `(du)/2`

= `int (du)/(2sqrt(u)) + int 4/sqrt((x - 2)^2 - (3)^2) dx`

= `1/2. sqrt(u)/(1/2) + 4 log |(x - 2) + sqrt((x - 2)^2 - (3)^2)| + C`

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

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2022-2023 (March) Outside Delhi Set 3

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