What is the integral of 3/(e^q+2) ?

The integral of 3/(e^q+2) is 3(1/2 q-1/2 ln|e^q+2|)+C

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integral of 3/(e^q+2)

\int \frac{3}{e ^{q} + 2} d q

3 (\frac{1}{2} q - \frac{1}{2} ln ∣ e^{q} + 2 ∣) + C

Solution steps

\int \frac{3}{e ^{q} + 2} d q

Take the constant out:

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

= 3 \cdot \int \frac{1}{e ^{q} + 2} d q

Apply u-substitution

: \int \frac{1}{u ( u + 2 )} d u

= 3 \cdot \int \frac{1}{u ( u + 2 )} d u

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= 3 \cdot \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} 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

= 3 (\int \frac{1}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u)

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= 3 (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣)

Substitute back

u = e^{q}

= 3 (\frac{1}{2} ln ∣ e^{q} ∣ - \frac{1}{2} ln ∣ e^{q} + 2 ∣)

\frac{1}{2} ln ∣ e^{q} ∣ = \frac{1}{2} q

= 3 (\frac{1}{2} q - \frac{1}{2} ln ∣ e^{q} + 2 ∣)

Add a constant to the solution

= 3 (\frac{1}{2} q - \frac{1}{2} ln ∣ e^{q} + 2 ∣) + C

\int e^{\frac{R}{L} t} d t

f (x) = 4 x - 3 x + 7

\int \frac{1}{x ^{2} + 2 x + 101} d x

\int \frac{2}{2 - x ^{2}} d x

What is the integral of 3/(e^q+2) ?
The integral of 3/(e^q+2) is 3(1/2 q-1/2 ln|e^q+2|)+C