The value of the integral _-1^2 0.05em log_e ⁡(x+ x^2+1) dx is

Question

Accepted Answer

Correct answer: D. (d) I = _-1^2 0.05em log_e ⁡(x+ x^2+1) dx =x log_e ⁡(x+ x^2+1)- _-1^2 0.05em ( 1+ x x^2+1x+ x^2+1 ) dx =x log_e ⁡(x+ x^2+1)- _-1^2 0.05em x x^2+1 dx =x log_e ⁡(x+ x^2+1)- ( x^2+1 )_-1^2 =(2 log_e ⁡ (2+ 5 )- 5) -(- log_e ⁡ (-1+ 2 )- 2) = log_e ⁡ (2+ 5 )^2- 5+ log_e ⁡ ( 2-1 )+ 2 = log_e ⁡ (2+ 5 )^2- 5+ log_e ⁡ ( 2-1 )+ 2 = 2- 5+ log_e ⁡ ( (2+ 5 )^2 2+1 ) = 2- 5+ log_e ⁡ ( 9+4 5 2+1 )

$The value of the integral \int_{- 1}^{2} log_{e} (x + x^{2} + 1) dx is$

Solution

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