What is the integral from 0 to ln(9) of tanh(x) ?

The integral from 0 to ln(9) of tanh(x) is ln(41/9)

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integral from 0 to ln(9) of tanh(x)

\int_{0}^{l n (9)} tanh (x) d x

ln (\frac{41}{9})

Decimal Notation

1.51634 \dots

Solution steps

\int_{0}^{l n (9)} tanh (x) d x

Rewrite using trig identities

= \int_{0}^{l n (9)} \frac{sinh ( x )}{cosh ( x )} d x

Apply u-substitution

: \int_{1}^{\frac{41}{9}} \frac{1}{u} d u

= \int_{1}^{\frac{41}{9}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{\frac{41}{9}}

Compute the boundaries:

ln (\frac{41}{9})

= ln (\frac{41}{9})

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What is the integral from 0 to ln(9) of tanh(x) ?
The integral from 0 to ln(9) of tanh(x) is ln(41/9)