What is the integral of 1/((10-2e^2x)) ?

The integral of 1/((10-2e^2x)) is -1/(2e^2)ln|10-2e^2x|+C

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integral of 1/((10-2e^2x))

integral

\frac{1}{( 10 - 2 e ^{2} x )}

- \frac{1}{2 e ^{2}} ln 10 - 2 e^{2} x + C

Solution steps

\int \frac{1}{10 - 2 e ^{2} x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2 e ^{2}} ln ∣ u ∣

Substitute back

u = 10 - 2 e^{2} x

= - \frac{1}{2 e ^{2}} ln 10 - 2 e^{2} x

Add a constant to the solution

= - \frac{1}{2 e ^{2}} ln 10 - 2 e^{2} x + C

\frac{1}{( 2 x - 3 ) ^{2}}

\frac{1}{( sin ( x ) cot ^{2} ( x ) )}

\frac{1}{( ( x - 3 ) ( x + 7 ) )}

- 5 cos (\frac{x}{2})

e^{(x^{2}) - x}

What is the integral of 1/((10-2e^2x)) ?
The integral of 1/((10-2e^2x)) is -1/(2e^2)ln|10-2e^2x|+C