What is the integral of ((e^7x))/((2+e^7x)) ?

The integral of ((e^7x))/((2+e^7x)) is 2/(e^7)+x-2/(e^7)ln|2+e^7x|+C

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integral of ((e^7x))/((2+e^7x))

integral

\frac{( e ^{7} x )}{( 2 + e ^{7} x )}

\frac{2}{e ^{7}} + x - \frac{2}{e ^{7}} ln 2 + e^{7} x + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 2}{e ^{14} u} d u

= e^{7} \cdot \int \frac{u - 2}{e ^{14} u} d u

Take the constant out:

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

= e^{7} \frac{1}{e ^{14}} \cdot \int \frac{u - 2}{u} d u

Expand

\frac{u - 2}{u} : 1 - \frac{2}{u}

= e^{7} \frac{1}{e ^{14}} \cdot \int 1 - \frac{2}{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

= e^{7} \frac{1}{e ^{14}} (\int 1 d u - \int \frac{2}{u} d u)

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= e^{7} \frac{1}{e ^{14}} (u - 2 ln ∣ u ∣)

Substitute back

u = 2 + e^{7} x

= e^{7} \frac{1}{e ^{14}} (2 + e^{7} x - 2 ln 2 + e^{7} x)

Simplify

e^{7} \frac{1}{e ^{14}} (2 + e^{7} x - 2 ln 2 + e^{7} x) : \frac{2}{e ^{7}} + x - \frac{2}{e ^{7}} ln 2 + e^{7} x

= \frac{2}{e ^{7}} + x - \frac{2}{e ^{7}} ln 2 + e^{7} x

Add a constant to the solution

= \frac{2}{e ^{7}} + x - \frac{2}{e ^{7}} ln 2 + e^{7} x + C

x^{2} - 8 x + 7

\frac{x}{( 1 + x ) ^{3}}

\frac{1}{( x ^{2} + a ^{2} ) ^{\frac{1}{2}}}

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

\frac{sin ^{4} ( x )}{cos ^{3} ( x )}

What is the integral of ((e^7x))/((2+e^7x)) ?
The integral of ((e^7x))/((2+e^7x)) is 2/(e^7)+x-2/(e^7)ln|2+e^7x|+C