What is the integral of (e^x(x-1)-e^x)/((x-1)^2) ?

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

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integral of (e^x(x-1)-e^x)/((x-1)^2)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u + 1} u - e ^{u + 1}}{u ^{2}} d u

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

Apply Integration By Parts

: - \frac{e ^{u + 1} u - e ^{u + 1}}{u} - \int - e^{u + 1} d u

= - \frac{e ^{u + 1} u - e ^{u + 1}}{u} - \int - e^{u + 1} d u

\int - e^{u + 1} d u = - e^{1 + u}

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

tan (\frac{1}{x})

\frac{1}{2 - x}, 1

\frac{d}{d x} (\frac{3 x ^{3} + x + 2}{5 x ^{2} + 1})

\frac{\partial}{\partial x} (7 ln (x^{2} + y^{2}))

n = 5 \sum \infty \frac{1}{n}

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