Question

Question #b418d

Answer 1

`int_1^e ((x^2+x+1)/x)dx=(e^2 + 2e - 1)/2`

Explanation:

`int_1^e ((x^2+x+1)/x)dx`

We know that `(x^2+x+1)/x = (x(x+1) + 1) / x = x+1+1/x`
So:

`int_1^e ((x^2+x+1)/x)dx = int_1^e (x+1+1/x)dx`

Using derivative rules we know that `int( x) dx = x^2/2 ; int (1) dx = x ; int(1/x) dx = ln(abs(x))`
To evaluate an integral you need the antiderivative
So:

#int_1^e (x+1+1/x)dx = |x^2/2 + x + ln(x)|_1^e = (e^2/2 + e + ln(e))- (1^2/2 + 1 + ln(1)) = (e^2+2e)/2 + 1 - 1/2 -1-ln(1) = (e^2 + 2e - 1)/2#