What is the integral from 1 to 4 of e^{-x}+(ln(x))/x ?

The integral from 1 to 4 of e^{-x}+(ln(x))/x is 1/e-1/(e^4)+2ln^2(2)

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integral from 1 to 4 of e^{-x}+(ln(x))/x

\int_{1}^{4} e^{- x} + \frac{ln ( x )}{x} d x

\frac{1}{e} - \frac{1}{e ^{4}} + 2 ln^{2} (2)

Decimal

1.31046 \dots

Solution steps

\int_{1}^{4} e^{- x} + \frac{ln ( x )}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{4} e^{- x} d x + \int_{1}^{4} \frac{ln ( x )}{x} d x

\int_{1}^{4} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{4}}

\int_{1}^{4} \frac{ln ( x )}{x} d x = 2 ln^{2} (2)

= \frac{1}{e} - \frac{1}{e ^{4}} + 2 ln^{2} (2)

What is the integral from 1 to 4 of e^{-x}+(ln(x))/x ?
The integral from 1 to 4 of e^{-x}+(ln(x))/x is 1/e-1/(e^4)+2ln^2(2)