What is the integral of (5(e^{2x}-x))/(x^2-e^{2x)} ?

The integral of (5(e^{2x}-x))/(x^2-e^{2x)} is -5/2 ln|x^2-e^{2x}|+C

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integral of (5(e^{2x}-x))/(x^2-e^{2x)}

\int \frac{5 ( e ^{2 x} - x )}{x ^{2} - e ^{2 x}} d x

- \frac{5}{2} ln x^{2} - e^{2 x} + C

Solution steps

\int \frac{5 ( e ^{2 x} - x )}{x ^{2} - e ^{2 x}} d x

Take the constant out:

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

= 5 \cdot \int \frac{e ^{2 x} - x}{x ^{2} - e ^{2 x}} d x

Apply u-substitution

: \int - \frac{1}{2 u} d u

= 5 \cdot \int - \frac{1}{2 u} d u

Take the constant out:

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

= 5 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = x^{2} - e^{2 x}

= 5 (- \frac{1}{2} ln x^{2} - e^{2 x})

Simplify

5 (- \frac{1}{2} ln x^{2} - e^{2 x}) : - \frac{5}{2} ln x^{2} - e^{2 x}

= - \frac{5}{2} ln x^{2} - e^{2 x}

Add a constant to the solution

= - \frac{5}{2} ln x^{2} - e^{2 x} + C

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What is the integral of (5(e^{2x}-x))/(x^2-e^{2x)} ?
The integral of (5(e^{2x}-x))/(x^2-e^{2x)} is -5/2 ln|x^2-e^{2x}|+C