What is the integral of (2sech^2(t))/(tanh(t)) ?

The integral of (2sech^2(t))/(tanh(t)) is 2(ln|sinh(t)|-ln|cosh(t)|)+C

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integral of (2sech^2(t))/(tanh(t))

\int \frac{2 sech ^{2} ( t )}{tanh ( t )} d t

2 (ln ∣ sinh (t) ∣ - ln ∣ cosh (t) ∣) + C

Solution steps

\int \frac{2 sech ^{2} ( t )}{tanh ( t )} d t

Take the constant out:

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

= 2 \cdot \int \frac{sech ^{2} ( t )}{tanh ( t )} d t

Rewrite using trig identities

= 2 \cdot \int \frac{1 - tanh ^{2} ( t )}{tanh ( t )} d t

Expand

\frac{1 - tanh ^{2} ( t )}{tanh ( t )} : \frac{1}{tanh ( t )} - tanh (t)

= 2 \cdot \int \frac{1}{tanh ( t )} - tanh (t) d t

Apply the Sum Rule:

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

= 2 (\int \frac{1}{tanh ( t )} d t - \int tanh (t) d t)

\int \frac{1}{tanh ( t )} d t = ln ∣ sinh (t) ∣

\int tanh (t) d t = ln ∣ cosh (t) ∣

= 2 (ln ∣ sinh (t) ∣ - ln ∣ cosh (t) ∣)

Add a constant to the solution

= 2 (ln ∣ sinh (t) ∣ - ln ∣ cosh (t) ∣) + C

n = 1 \sum \infty \frac{n}{4 ( n + 1 ) !}

ln (\frac{x ^{2} ( x + 1 )}{x + 2})

\int \frac{5 + 4 x}{1 + x ^{2}} d x

\int e^{- \frac{x}{8}} d x

x^{^{''}} + 2 x^{^{'}} + 2 = 0, x (0) = 0, x^{^{'}} (0) = 1

What is the integral of (2sech^2(t))/(tanh(t)) ?
The integral of (2sech^2(t))/(tanh(t)) is 2(ln|sinh(t)|-ln|cosh(t)|)+C