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∫1x2+4x+8dx\int \frac{1}{x^2+4x+8}dx is equal to (Where C is an arbitrary constant)

Solution

✅ Correct Option: 4

Complete the square: x2+4x+8=(x+2)2+22x^2+4x+8 = (x+2)^2+2^2.

Using ∫dtt2+a2=1atan⁡−1ta\int\frac{dt}{t^2+a^2} = \frac{1}{a}\tan^{-1}\frac{t}{a}, the integral is 12tan⁡−1x+22+C\frac{1}{2}\tan^{-1}\frac{x+2}{2}+C.

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