What is the integral of (e^{ax+b}-e^{bx+a})/(e^{ax)} ?

The integral of (e^{ax+b}-e^{bx+a})/(e^{ax)} is e^bx+b/(ae^{-b)}-(e^{-ax+bx+a})/(b-a)+C

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integral of (e^{ax+b}-e^{bx+a})/(e^{ax)}

\int \frac{e ^{a x + b} - e ^{b x + a}}{e ^{a x}} d x

e^{b} x + \frac{b}{a e ^{- b}} - \frac{e ^{- a x + b x + a}}{b - a} + C

Solution steps

\int \frac{e ^{a x + b} - e ^{b x + a}}{e ^{a x}} d x

Apply u-substitution

= \int \frac{u - e ^{\frac{b ( l n ( u ) - b )}{a} + a}}{a e ^{- b} u ^{2}} d u

Take the constant out:

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

= \frac{1}{a e ^{- b}} \cdot \int \frac{u - e ^{\frac{b ( l n ( u ) - b )}{a} + a}}{u ^{2}} d u

Expand

\frac{u - e ^{\frac{b ( l n ( u ) - b )}{a} + a}}{u ^{2}} : \frac{1}{u} - e^{- \frac{b ^{2}}{a} + a} u^{\frac{b - 2 a}{a}}

= \frac{1}{a e ^{- b}} \cdot \int \frac{1}{u} - e^{- \frac{b ^{2}}{a} + a} u^{\frac{b - 2 a}{a}} 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

= \frac{1}{a e ^{- b}} (\int \frac{1}{u} d u - \int e^{- \frac{b ^{2}}{a} + a} u^{\frac{b - 2 a}{a}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int e^{- \frac{b ^{2}}{a} + a} u^{\frac{b - 2 a}{a}} d u = \frac{a e ^{\frac{- b ^{2} + a ^{2}}{a}} u ^{\frac{b - a}{a}}}{b - a}

= \frac{1}{a e ^{- b}} (ln ∣ u ∣ - \frac{a e ^{\frac{- b ^{2} + a ^{2}}{a}} u ^{\frac{b - a}{a}}}{b - a})

Substitute back

u = e^{a x + b}

= \frac{1}{a e ^{- b}} ln e^{a x + b} - \frac{a e ^{\frac{- b ^{2} + a ^{2}}{a}} ( e ^{a x + b} ) ^{\frac{b - a}{a}}}{b - a}

Simplify

\frac{1}{a e ^{- b}} ln e^{a x + b} - \frac{a e ^{\frac{- b ^{2} + a ^{2}}{a}} ( e ^{a x + b} ) ^{\frac{b - a}{a}}}{b - a} : e^{b} x + \frac{b}{a e ^{- b}} - \frac{e ^{- a x + b x + a}}{b - a}

= e^{b} x + \frac{b}{a e ^{- b}} - \frac{e ^{- a x + b x + a}}{b - a}

Add a constant to the solution

= e^{b} x + \frac{b}{a e ^{- b}} - \frac{e ^{- a x + b x + a}}{b - a} + C

What is the integral of (e^{ax+b}-e^{bx+a})/(e^{ax)} ?
The integral of (e^{ax+b}-e^{bx+a})/(e^{ax)} is e^bx+b/(ae^{-b)}-(e^{-ax+bx+a})/(b-a)+C