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प्रश्न
Find `int dx/(x^2 + 4x + 8)`
उत्तर
Consider the integral, `int dx/(x^2 + 4x + 8)`
I = `int 1/(x^2 + 4x + 8)` dx
= `int 1/(x^2 + 4x + 4 + 4)` dx
= `int 1/((x + 2)^2 + 2^2) dx` {Use Intergral Formula : `int 1/(x^2 + a^2) dx = 1/a tan^(-1) (x/a)}`
`= 1/2 tan^(-1) (x + 2)/2 + C`
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