What is the integral from 1 to e of (sec(2+ln(x)))/x ?

The integral from 1 to e of (sec(2+ln(x)))/x is ln(-tan(3)-sec(3))-ln(-tan(2)-sec(2))

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integral from 1 to e of (sec(2+ln(x)))/x

\int_{1}^{e} \frac{sec ( 2 + ln ( x ) )}{x} d x

ln (- tan (3) - sec (3)) - ln (- tan (2) - sec (2))

Decimal Notation

- 1.38138 \dots

Solution steps

\int_{1}^{e} \frac{sec ( 2 + ln ( x ) )}{x} d x

Apply u-substitution

: \int_{2}^{3} sec (u) d u

= \int_{2}^{3} sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= [ln ∣ tan (u) + sec (u) ∣]_{2}^{3}

Compute the boundaries:

ln (- tan (3) - sec (3)) - ln (- tan (2) - sec (2))

= ln (- tan (3) - sec (3)) - ln (- tan (2) - sec (2))

\int e^{2 x + l n (x)} d x

x \to 7 - lim (\frac{1}{x ^{2} ( x + 7 )})

\int \frac{2 x ^{3}}{2 x ^{2} - 4 x + 3} d x

\int x^{3} d x

\int 6 x e^{x^{2}} d x

What is the integral from 1 to e of (sec(2+ln(x)))/x ?
The integral from 1 to e of (sec(2+ln(x)))/x is ln(-tan(3)-sec(3))-ln(-tan(2)-sec(2))