What is the integral of (2e^{2r})/(7+e^{2r)} ?

The integral of (2e^{2r})/(7+e^{2r)} is ln|7+e^{2r}|+C

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integral of (2e^{2r})/(7+e^{2r)}

\int \frac{2 e ^{2 r}}{7 + e ^{2 r}} d r

ln 7 + e^{2 r} + C

Solution steps

\int \frac{2 e ^{2 r}}{7 + e ^{2 r}} d r

Take the constant out:

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

= 2 \cdot \int \frac{e ^{2 r}}{7 + e ^{2 r}} d r

Apply u-substitution

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

= 2 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 7 + e^{2 r}

= 2 \cdot \frac{1}{2} ln 7 + e^{2 r}

Simplify

2 \cdot \frac{1}{2} ln 7 + e^{2 r} : ln 7 + e^{2 r}

= ln 7 + e^{2 r}

Add a constant to the solution

= ln 7 + e^{2 r} + C

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What is the integral of (2e^{2r})/(7+e^{2r)} ?
The integral of (2e^{2r})/(7+e^{2r)} is ln|7+e^{2r}|+C