`int_1^4(1/(e^x+pi))dx`

Lets first integrate the indefinite integral,

`int(1/(e^x+pi))dx`

let `u = e^x+pi`

then, `du = e^x dx`

therefore, `dx = (du)/e^x`

`e^x =u-pi`

substituting,

`= int1/(u)*((du)/(u-pi))`

by separating into fractions,

`=int-1/(piu) du + int1/(pi(u-pi))du`

`=(-1/pi)ln(u)+ (1/pi)ln(u-pi) `

`=(-1/pi)ln(e^x+pi) + (1/pi)ln(e^x+pi-pi)`

`=-ln(e^x+pi)/pi +x/pi`

`int1/(e^x+pi)dx=x/pi - ln(e^x+pi)/pi`

therefore, the definite integral,

`int_1^4 1/(e^x+pi)dx = (4-1)/pi - (ln(e^4+pi)-ln(e^1+pi))/pi`

`= (3 - ln((e^4+pi)/(e+pi)))/pi`

= 0.226694719