Integrate (x²+1)/[(x+1) (x+2)] dx

Answer 1

We can rewrite the integrand as:

(x²+1)/[(x+1) (x+2)] = A/(x+1) + B/(x+2)

where A and B are constants to be determined. Multiplying both sides by (x+1) (x+2), we get:

x²+1 = A(x+2) + B(x+1)

Expanding and equating coefficients of x², x and constants, we get:

A + B = 0
2A + B = 1
A = 1/3, B = -1/3

Therefore, we have:

(x²+1)/[(x+1) (x+2)] = 1/3/(x+1) - 1/3/(x+2)

Integrating both sides, we get:

∫(x²+1)/[(x+1) (x+2)] dx = ∫1/3/(x+1) dx - ∫1/3/(x+2) dx

= (1/3) ln|x+1| - (1/3) ln|x+2| + C

where C is the constant of integration.

Answer 2

AAAaannndd the bot gets it wrong yet again!
actually,
(x^2+1)/[(x+1) (x+2)] = 1 + 2/(x+1) - 5/(x+2)
so the integral is just
x + 2ln(x+1) - 5ln(x+2) + C

Answer 3

You are correct, I apologize for the mistake. Thank you for the correction.

The correct partial fraction decomposition is indeed:

(x^2+1)/[(x+1) (x+2)] = 1 + 2/(x+1) - 5/(x+2)

And the integral is then simply:

∫(x²+1)/[(x+1) (x+2)] dx = ∫[1 + 2/(x+1) - 5/(x+2)] dx

= x + 2 ln|x+1| - 5 ln|x+2| + C

where C is the constant of integration.

Thank you again for correcting me.