What is the integral of (e^{4x}+4e^x+8)/(e^x) ?

The integral of (e^{4x}+4e^x+8)/(e^x) is -(e^{4x}+4e^x+8)/(e^x)+4(1/3 e^{3x}+x)+C

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integral of (e^{4x}+4e^x+8)/(e^x)

\int \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} d x

- \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} + 4 (\frac{1}{3} e^{3 x} + x) + C

Solution steps

\int \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} d x

Apply Integration By Parts

: - \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} - \int - 4 (e^{3 x} + 1) d x

= - \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} - \int - 4 (e^{3 x} + 1) d x

\int - 4 (e^{3 x} + 1) d x = - 4 (\frac{1}{3} e^{3 x} + x)

= - \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} - (- 4 (\frac{1}{3} e^{3 x} + x))

Simplify

= - \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} + 4 (\frac{1}{3} e^{3 x} + x)

Add a constant to the solution

= - \frac{e ^{4 x} + 4 e ^{x} + 8}{e ^{x}} + 4 (\frac{1}{3} e^{3 x} + x) + C

\frac{d}{d x} (cos^{2} (1 - x))

\frac{2}{s ^{2} + 40 s + 400}

\int_{0}^{4} e^{2 x} d x

(a x^{2} + b x + c) \cdot e^{- \frac{x}{2}}

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

What is the integral of (e^{4x}+4e^x+8)/(e^x) ?
The integral of (e^{4x}+4e^x+8)/(e^x) is -(e^{4x}+4e^x+8)/(e^x)+4(1/3 e^{3x}+x)+C