What is the integral of (5-e^x)/(e^{2x)} ?

The integral of (5-e^x)/(e^{2x)} is -5/(2e^{2x)}+1/(e^x)+C

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integral of (5-e^x)/(e^{2x)}

\int \frac{5 - e ^{x}}{e ^{2 x}} d x

- \frac{5}{2 e ^{2 x}} + \frac{1}{e ^{x}} + C

Solution steps

\int \frac{5 - e ^{x}}{e ^{2 x}} d x

Apply u-substitution

: \int \frac{5 - u}{u ^{3}} d u

= \int \frac{5 - u}{u ^{3}} d u

Expand

\frac{5 - u}{u ^{3}} : \frac{5}{u ^{3}} - \frac{1}{u ^{2}}

= \int \frac{5}{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{5}{u ^{3}} d u - \int \frac{1}{u ^{2}} d u

\int \frac{5}{u ^{3}} d u = - \frac{5}{2 u ^{2}}

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

= - \frac{5}{2 u ^{2}} - (- \frac{1}{u})

Substitute back

u = e^{x}

= - \frac{5}{2 ( e ^{x} ) ^{2}} - (- \frac{1}{e ^{x}})

Simplify

- \frac{5}{2 ( e ^{x} ) ^{2}} - (- \frac{1}{e ^{x}}) : - \frac{5}{2 e ^{2 x}} + \frac{1}{e ^{x}}

= - \frac{5}{2 e ^{2 x}} + \frac{1}{e ^{x}}

Add a constant to the solution

= - \frac{5}{2 e ^{2 x}} + \frac{1}{e ^{x}} + C

\int \frac{1}{( 64 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{220 x}{( x + 2 ) ^{2}}

\int (1 - \frac{3}{y} + x) d y

\int_{1}^{4} (x^{3} + \frac{1}{4 x ^{3}}) d x

n = 0 \sum \infty \frac{n ^{6}}{2 ^{n}}

What is the integral of (5-e^x)/(e^{2x)} ?
The integral of (5-e^x)/(e^{2x)} is -5/(2e^{2x)}+1/(e^x)+C