What is the integral of (csch(1))/2 x(coth(1))/2 x ?

The integral of (csch(1))/2 x(coth(1))/2 x is ((e^3+e)x^3)/(6(e^2-1)^2)+C

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integral of (csch(1))/2 x(coth(1))/2 x

\int \frac{csch ( 1 )}{2} x \frac{coth ( 1 )}{2} x d x

\frac{( e ^{3} + e ) x ^{3}}{6 ( e ^{2} - 1 ) ^{2}} + C

Solution steps

\int \frac{csch ( 1 )}{2} x \frac{coth ( 1 )}{2} x d x

Take the constant out:

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

= \frac{csch ( 1 )}{2} \cdot \frac{coth ( 1 )}{2} \cdot \int xx d x

\frac{csch ( 1 )}{2} \cdot \frac{coth ( 1 )}{2} = \frac{e ^{3} + e}{2 ( e ^{2} - 1 ) ^{2}}

= \frac{e ^{3} + e}{2 ( e ^{2} - 1 ) ^{2}} \cdot \int xx d x

Simplify

xx : x^{2}

= \frac{e ^{3} + e}{2 ( e ^{2} - 1 ) ^{2}} \cdot \int x^{2} d x

Apply the Power Rule

: \frac{x ^{3}}{3}

= \frac{e ^{3} + e}{2 ( e ^{2} - 1 ) ^{2}} \cdot \frac{x ^{3}}{3}

Simplify

\frac{e ^{3} + e}{2 ( e ^{2} - 1 ) ^{2}} \cdot \frac{x ^{3}}{3} : \frac{( e ^{3} + e ) x ^{3}}{6 ( e ^{2} - 1 ) ^{2}}

= \frac{( e ^{3} + e ) x ^{3}}{6 ( e ^{2} - 1 ) ^{2}}

Add a constant to the solution

= \frac{( e ^{3} + e ) x ^{3}}{6 ( e ^{2} - 1 ) ^{2}} + C

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What is the integral of (csch(1))/2 x(coth(1))/2 x ?
The integral of (csch(1))/2 x(coth(1))/2 x is ((e^3+e)x^3)/(6(e^2-1)^2)+C