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

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

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integral of (e^{4x}-e^{2x}+1)/(e^x)

\int \frac{e ^{4 x} - e ^{2 x} + 1}{e ^{x}} d x

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

Solution steps

\int \frac{e ^{4 x} - e ^{2 x} + 1}{e ^{x}} d x

Apply u-substitution

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

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

Expand

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

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

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

\int 1 d u = u

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

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

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = y

f (x) = 4 x - 6

tan (sec (cos (t)))

\frac{d}{d x} (x^{0})

x \to 5 lim (4 x^{3} - 6 x^{2})

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