What is the integral of (5-e^x)/(e^{6x)} ?

The integral of (5-e^x)/(e^{6x)} is -5/(6e^{6x)}+1/(5e^{5x)}+C

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integral of (5-e^x)/(e^{6x)}

\int \frac{5 - e ^{x}}{e ^{6 x}} d x

- \frac{5}{6 e ^{6 x}} + \frac{1}{5 e ^{5 x}} + C

Solution steps

\int \frac{5 - e ^{x}}{e ^{6 x}} d x

Apply u-substitution

: \int \frac{5 - u}{u ^{7}} d u

= \int \frac{5 - u}{u ^{7}} d u

Expand

\frac{5 - u}{u ^{7}} : \frac{5}{u ^{7}} - \frac{1}{u ^{6}}

= \int \frac{5}{u ^{7}} - \frac{1}{u ^{6}} 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 ^{7}} d u - \int \frac{1}{u ^{6}} d u

\int \frac{5}{u ^{7}} d u = - \frac{5}{6 u ^{6}}

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

= - \frac{5}{6 u ^{6}} - (- \frac{1}{5 u ^{5}})

Substitute back

u = e^{x}

= - \frac{5}{6 ( e ^{x} ) ^{6}} - (- \frac{1}{5 ( e ^{x} ) ^{5}})

Simplify

- \frac{5}{6 ( e ^{x} ) ^{6}} - (- \frac{1}{5 ( e ^{x} ) ^{5}}) : - \frac{5}{6 e ^{6 x}} + \frac{1}{5 e ^{5 x}}

= - \frac{5}{6 e ^{6 x}} + \frac{1}{5 e ^{5 x}}

Add a constant to the solution

= - \frac{5}{6 e ^{6 x}} + \frac{1}{5 e ^{5 x}} + C

\int x^{2} (x^{3} + 5)^{10} d x

y^{^{''}} + 2 y^{^{'}} + y = 3 - e^{- t}

\frac{d}{d x} (sec^{2} (x) + tan^{2} (x))

\int cos^{8} (x) d x

(6.4) (2.7)

What is the integral of (5-e^x)/(e^{6x)} ?
The integral of (5-e^x)/(e^{6x)} is -5/(6e^{6x)}+1/(5e^{5x)}+C