What is the integral of (e^x(e^x+1))/((e^x-1)^3) ?

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

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integral of (e^x(e^x+1))/((e^x-1)^3)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 1}{( u - 1 ) ^{3}} d u

= \int \frac{u + 1}{( u - 1 ) ^{3}} d u

Expand

\frac{u + 1}{( u - 1 ) ^{3}} : \frac{u}{( u - 1 ) ^{3}} + \frac{1}{( u - 1 ) ^{3}}

= \int \frac{u}{( u - 1 ) ^{3}} + \frac{1}{( u - 1 ) ^{3}} d u

Apply the Sum Rule:

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

= \int \frac{u}{( u - 1 ) ^{3}} d u + \int \frac{1}{( u - 1 ) ^{3}} d u

\int \frac{u}{( u - 1 ) ^{3}} d u = - \frac{1}{u - 1} - \frac{1}{2 ( u - 1 ) ^{2}}

\int \frac{1}{( u - 1 ) ^{3}} d u = - \frac{1}{2 ( u - 1 ) ^{2}}

= - \frac{1}{u - 1} - \frac{1}{2 ( u - 1 ) ^{2}} - \frac{1}{2 ( u - 1 ) ^{2}}

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = 3 (2 - y)

\frac{d y}{d x} = \frac{x y + y ^{2}}{x ^{2}}

\frac{d y}{d x} = 16 4 4 x^{4} + 4

y^{^{''}} + 2 y^{^{'}} + 26 y = 0

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

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