What is the integral of (e^x+e^{2x}+e^{3x})/(e^{4x)} ?

The integral of (e^x+e^{2x}+e^{3x})/(e^{4x)} is -1/(3e^{3x)}-1/(2e^{2x)}-1/(e^x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (e^x+e^{2x}+e^{3x})/(e^{4x)}

\int \frac{e ^{x} + e ^{2 x} + e ^{3 x}}{e ^{4 x}} d x

- \frac{1}{3 e ^{3 x}} - \frac{1}{2 e ^{2 x}} - \frac{1}{e ^{x}} + C

Solution steps

\int \frac{e ^{x} + e ^{2 x} + e ^{3 x}}{e ^{4 x}} d x

Apply u-substitution

= \int \frac{1 + u + u ^{2}}{u ^{4}} d u

Expand

\frac{1 + u + u ^{2}}{u ^{4}} : \frac{1}{u ^{4}} + \frac{1}{u ^{3}} + \frac{1}{u ^{2}}

= \int \frac{1}{u ^{4}} + \frac{1}{u ^{3}} + \frac{1}{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

= \int \frac{1}{u ^{4}} d u + \int \frac{1}{u ^{3}} d u + \int \frac{1}{u ^{2}} d u

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= - \frac{1}{3 u ^{3}} - \frac{1}{2 u ^{2}} - \frac{1}{u}

Substitute back

u = e^{x}

= - \frac{1}{3 ( e ^{x} ) ^{3}} - \frac{1}{2 ( e ^{x} ) ^{2}} - \frac{1}{e ^{x}}

Simplify

- \frac{1}{3 ( e ^{x} ) ^{3}} - \frac{1}{2 ( e ^{x} ) ^{2}} - \frac{1}{e ^{x}} : - \frac{1}{3 e ^{3 x}} - \frac{1}{2 e ^{2 x}} - \frac{1}{e ^{x}}

= - \frac{1}{3 e ^{3 x}} - \frac{1}{2 e ^{2 x}} - \frac{1}{e ^{x}}

Add a constant to the solution

= - \frac{1}{3 e ^{3 x}} - \frac{1}{2 e ^{2 x}} - \frac{1}{e ^{x}} + C

\int \frac{x ^{2}}{3 ( 1 + x ^{2} )} d x

\int x 7 x^{2} + 5 d x

\int \frac{1}{sin ^{2} ( x ) + 5 cos ^{2} ( x )} d x

\int \frac{1}{( 8 + x ) ^{2}} d x

\int \frac{x ^{3}}{4 - x ^{8}} d x

What is the integral of (e^x+e^{2x}+e^{3x})/(e^{4x)} ?
The integral of (e^x+e^{2x}+e^{3x})/(e^{4x)} is -1/(3e^{3x)}-1/(2e^{2x)}-1/(e^x)+C