What is the integral of 1/(1+cosh(x)) ?

The integral of 1/(1+cosh(x)) is -2/(e^x+1)+C

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integral of 1/(1+cosh(x))

\int \frac{1}{1 + cosh ( x )} d x

- \frac{2}{e ^{x} + 1} + C

Solution steps

\int \frac{1}{1 + cosh ( x )} d x

Rewrite using trig identities

= \int \frac{2}{2 + e ^{- x} + e ^{x}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{2 + e ^{- x} + e ^{x}} d x

Apply u-substitution

= 2 \cdot \int \frac{1}{2 u + 1 + u ^{2}} d u

Complete the square

2 u + 1 + u^{2} : (u + 1)^{2}

= 2 \cdot \int \frac{1}{( u + 1 ) ^{2}} d u

Apply u-substitution

= 2 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

= 2 (- \frac{1}{v})

Substitute back

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

Simplify

2 (- \frac{1}{e ^{x} + 1}) : - \frac{2}{e ^{x} + 1}

= - \frac{2}{e ^{x} + 1}

Add a constant to the solution

= - \frac{2}{e ^{x} + 1} + C

What is the integral of 1/(1+cosh(x)) ?
The integral of 1/(1+cosh(x)) is -2/(e^x+1)+C