Solve ∫ 1/e^lnx+C-1 wrt y

$$\int \frac { 1 } { e ^ { \ln x + C } - 1 } d y$$

Evaluate

$$\frac{y}{xe^{C}-1}+С$$

$$x\neq \frac{1}{e^{C}}$$

Solution Steps

Find the integral of \frac{1}{e^{\log_{e}\left(x\right)+C}-1} using the table of common integrals rule \int a\mathrm{d}y=ay.

$$\frac{y}{e^{\log_{e}\left(x\right)+C}-1}$$

Simplify.

$$\frac{y}{xe^{C}-1}$$

If F\left(y\right) is an antiderivative of f\left(y\right), then the set of all antiderivatives of f\left(y\right) is given by F\left(y\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\begin{matrix}\frac{y}{xe^{C}-1}+С_{3},&\end{matrix}$$

Differentiate w.r.t. x

$$-\frac{ye^{C}}{\left(xe^{C}-1\right)^{2}}$$