What is the integral of (e^x(2x^3+x^2-4x-6))/(6x^4) ?

The integral of (e^x(2x^3+x^2-4x-6))/(6x^4) is (2e^xx^2+3e^xx+2e^x)/(6x^3)+C

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integral of (e^x(2x^3+x^2-4x-6))/(6x^4)

\int \frac{e ^{x} ( 2 x ^{3} + x ^{2} - 4 x - 6 )}{6 x ^{4}} d x

\frac{2 e ^{x} x ^{2} + 3 e ^{x} x + 2 e ^{x}}{6 x ^{3}} + C

Solution steps

\int \frac{e ^{x} ( 2 x ^{3} + x ^{2} - 4 x - 6 )}{6 x ^{4}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{e ^{x} ( 2 x ^{3} + x ^{2} - 4 x - 6 )}{x ^{4}} d x

Expand

\frac{e ^{x} ( 2 x ^{3} + x ^{2} - 4 x - 6 )}{x ^{4}} : \frac{2 e ^{x}}{x} + \frac{e ^{x}}{x ^{2}} - \frac{4 e ^{x}}{x ^{3}} - \frac{6 e ^{x}}{x ^{4}}

= \frac{1}{6} \cdot \int \frac{2 e ^{x}}{x} + \frac{e ^{x}}{x ^{2}} - \frac{4 e ^{x}}{x ^{3}} - \frac{6 e ^{x}}{x ^{4}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (\int \frac{2 e ^{x}}{x} d x + \int \frac{e ^{x}}{x ^{2}} d x - \int \frac{4 e ^{x}}{x ^{3}} d x - \int \frac{6 e ^{x}}{x ^{4}} d x)

\int \frac{2 e ^{x}}{x} d x = 2 Ei (x)

\int \frac{e ^{x}}{x ^{2}} d x = - \frac{e ^{x}}{x} + Ei (x)

\int \frac{4 e ^{x}}{x ^{3}} d x = 4 (- \frac{e ^{x}}{2 x ^{2}} + \frac{1}{2} (- \frac{e ^{x}}{x} + Ei (x)))

\int \frac{6 e ^{x}}{x ^{4}} d x = 6 (- \frac{e ^{x}}{3 x ^{3}} + \frac{1}{3} (- \frac{e ^{x}}{2 x ^{2}} + \frac{1}{2} (- \frac{e ^{x}}{x} + Ei (x))))

= \frac{1}{6} (2 Ei (x) - \frac{e ^{x}}{x} + Ei (x) - 4 (- \frac{e ^{x}}{2 x ^{2}} + \frac{1}{2} (- \frac{e ^{x}}{x} + Ei (x))) - 6 (- \frac{e ^{x}}{3 x ^{3}} + \frac{1}{3} (- \frac{e ^{x}}{2 x ^{2}} + \frac{1}{2} (- \frac{e ^{x}}{x} + Ei (x)))))

Simplify

\frac{1}{6} (2 Ei (x) - \frac{e ^{x}}{x} + Ei (x) - 4 (- \frac{e ^{x}}{2 x ^{2}} + \frac{1}{2} (- \frac{e ^{x}}{x} + Ei (x))) - 6 (- \frac{e ^{x}}{3 x ^{3}} + \frac{1}{3} (- \frac{e ^{x}}{2 x ^{2}} + \frac{1}{2} (- \frac{e ^{x}}{x} + Ei (x))))) : \frac{2 e ^{x} x ^{2} + 3 e ^{x} x + 2 e ^{x}}{6 x ^{3}}

= \frac{2 e ^{x} x ^{2} + 3 e ^{x} x + 2 e ^{x}}{6 x ^{3}}

Add a constant to the solution

= \frac{2 e ^{x} x ^{2} + 3 e ^{x} x + 2 e ^{x}}{6 x ^{3}} + C

t \to 0 lim (\frac{sin ( 2 t )}{t})

x \to 0 + lim (sin (x) \cdot ln (x))

4 y^{^{''}} - 9 y = 0

y = \frac{1}{t + 1}

\frac{d}{d x} (\frac{1}{4} x^{3} - 2 x)

What is the integral of (e^x(2x^3+x^2-4x-6))/(6x^4) ?
The integral of (e^x(2x^3+x^2-4x-6))/(6x^4) is (2e^xx^2+3e^xx+2e^x)/(6x^3)+C