What is the integral of (e^x)/(e^4-1) ?

The integral of (e^x)/(e^4-1) is 1/(e^4-1)e^x+C

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integral of (e^x)/(e^4-1)

\int \frac{e ^{x}}{e ^{4} - 1} d x

\frac{1}{e ^{4} - 1} e^{x} + C

Solution steps

\int \frac{e ^{x}}{e ^{4} - 1} d x

Take the constant out:

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

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

Use the common integral:

\int e^{x} d x = e^{x}

= \frac{1}{e ^{4} - 1} e^{x}

Add a constant to the solution

= \frac{1}{e ^{4} - 1} e^{x} + C

\frac{d y}{d x} = 8 y, y (0) = 6

\int \frac{x ^{2} - 29 x + 5}{( x - 4 ) ^{2} ( x ^{2} + 3 )} d x

\int_{0}^{4} 1 + 9 x d x

ln (1 - 3 y)

\int y cos (y^{2}) d y

What is the integral of (e^x)/(e^4-1) ?
The integral of (e^x)/(e^4-1) is 1/(e^4-1)e^x+C