What is the integral from 1 to e of cosh(ln(x)) ?

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

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integral from 1 to e of cosh(ln(x))

\int_{1}^{e} cosh (ln (x)) d x

\frac{e ^{2} + 1}{4}

Decimal Notation

2.09726 \dots

Solution steps

\int_{1}^{e} cosh (ln (x)) d x

Apply Integration By Parts

: [x cosh (ln (x)) - \int sinh (ln (x)) d x]_{1}^{e}

= [x cosh (ln (x)) - \int sinh (ln (x)) d x]_{1}^{e}

\int sinh (ln (x)) d x = x sinh (ln (x)) - \frac{1}{2} (ln ∣ x ∣ + \frac{x ^{2}}{2})

= [x cosh (ln (x)) - (x sinh (ln (x)) - \frac{1}{2} (ln ∣ x ∣ + \frac{x ^{2}}{2}))]_{1}^{e}

Simplify

[x cosh (ln (x)) - (x sinh (ln (x)) - \frac{1}{2} (ln ∣ x ∣ + \frac{x ^{2}}{2}))]_{1}^{e} : [\frac{1}{4} (4 x cosh (ln (x)) - 4 x sinh (ln (x)) + 2 ln ∣ x ∣ + x^{2})]_{1}^{e}

= [\frac{1}{4} (4 x cosh (ln (x)) - 4 x sinh (ln (x)) + 2 ln ∣ x ∣ + x^{2})]_{1}^{e}

Compute the boundaries:

\frac{e ^{2} + 6}{4} - \frac{5}{4}

= \frac{e ^{2} + 6}{4} - \frac{5}{4}

Simplify

= \frac{e ^{2} + 1}{4}

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