Solve ∫ e^2x-1/e^x+1 wrt x=?

$$\int \frac { e ^ { 2 x } - 1 } { e ^ { x } + 1 } d x = ?$$

Evaluate

$$С+e^{x}-x$$

Solution Steps

Factor the expressions that are not already factored in \frac{e^{2x}-1}{e^{x}+1}.

$$\int \frac{\left(e^{x}-1\right)\left(e^{x}+1\right)}{e^{x}+1}\mathrm{d}x$$

Cancel out e^{x}+1 in both numerator and denominator.

$$\int e^{x}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int e^{x}\mathrm{d}x+\int -1\mathrm{d}x$$

Use \int e^{x}\mathrm{d}x=e^{x} from the table of common integrals to obtain the result.

$$e^{x}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$e^{x}-x$$

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

$$e^{x}-x+С$$

Differentiate w.r.t. x

$$e^{x}-1$$