What is the integral of 4/(4-5^x) ?

The integral of 4/(4-5^x) is -(-ln(5)x+ln|4-5^x|)/(ln(5))+C

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integral of 4/(4-5^x)

\int \frac{4}{4 - 5 ^{x}} d x

- \frac{- ln ( 5 ) x + ln ∣ 4 - 5 ^{x} ∣}{ln ( 5 )} + C

Solution steps

\int \frac{4}{4 - 5 ^{x}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 - 5 ^{x}} d x

Apply u-substitution

: \int - \frac{1}{ln ( 5 ) ( - u + 4 ) u} d u

= 4 \cdot \int - \frac{1}{ln ( 5 ) ( - u + 4 ) u} d u

Take the constant out:

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

= 4 (- \frac{1}{ln ( 5 )} \cdot \int \frac{1}{( - u + 4 ) u} d u)

Take the partial fraction of

\frac{1}{( - u + 4 ) u} : - \frac{1}{4 ( u - 4 )} + \frac{1}{4 u}

= 4 (- \frac{1}{ln ( 5 )} \cdot \int - \frac{1}{4 ( u - 4 )} + \frac{1}{4 u} d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (- \frac{1}{ln ( 5 )} (- \int \frac{1}{4 ( u - 4 )} d u + \int \frac{1}{4 u} d u))

\int \frac{1}{4 ( u - 4 )} d u = \frac{1}{4} ln ∣ u - 4 ∣

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

= 4 (- \frac{1}{ln ( 5 )} (- \frac{1}{4} ln ∣ u - 4 ∣ + \frac{1}{4} ln ∣ u ∣))

Substitute back

u = 4 - 5^{x}

= 4 (- \frac{1}{ln ( 5 )} (- \frac{1}{4} ln ∣ 4 - 5^{x} - 4 ∣ + \frac{1}{4} ln ∣ 4 - 5^{x} ∣))

Simplify

4 (- \frac{1}{ln ( 5 )} (- \frac{1}{4} ln ∣ 4 - 5^{x} - 4 ∣ + \frac{1}{4} ln ∣ 4 - 5^{x} ∣)) : - \frac{- ln ( 5 ) x + ln ∣ 4 - 5 ^{x} ∣}{ln ( 5 )}

= - \frac{- ln ( 5 ) x + ln ∣ 4 - 5 ^{x} ∣}{ln ( 5 )}

Add a constant to the solution

= - \frac{- ln ( 5 ) x + ln ∣ 4 - 5 ^{x} ∣}{ln ( 5 )} + C

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What is the integral of 4/(4-5^x) ?
The integral of 4/(4-5^x) is -(-ln(5)x+ln|4-5^x|)/(ln(5))+C