What is the integral of 3^x+1/x-x^{sqrt(3)} ?

The integral of 3^x+1/x-x^{sqrt(3)} is (3^x}{ln(3)}+ln|x|-\frac{sqrt(3)-1)/2 x^{sqrt(3)+1}+C

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integral of 3^x+1/x-x^{sqrt(3)}

\int 3^{x} + \frac{1}{x} - x^{3} d x

\frac{3 ^{x}}{ln ( 3 )} + ln ∣ x ∣ - \frac{3 - 1}{2} x^{3 + 1} + C

Solution steps

\int 3^{x} + \frac{1}{x} - x^{3} 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

= \int 3^{x} d x + \int \frac{1}{x} d x - \int x^{3} d x

\int 3^{x} d x = \frac{3 ^{x}}{ln ( 3 )}

\int \frac{1}{x} d x = ln ∣ x ∣

\int x^{3} d x = \frac{3 - 1}{2} x^{3 + 1}

= \frac{3 ^{x}}{ln ( 3 )} + ln ∣ x ∣ - \frac{3 - 1}{2} x^{3 + 1}

Add a constant to the solution

= \frac{3 ^{x}}{ln ( 3 )} + ln ∣ x ∣ - \frac{3 - 1}{2} x^{3 + 1} + C

\frac{d}{d x} (\frac{8 x}{x ^{2} + 3})

\frac{d}{d x} [f (x^{2})] = x^{5}

\frac{\partial}{\partial x} (x^{2} + 2 x y + 3 y^{4})

\int \frac{1}{x x + 1} d x

y = ln (\frac{4}{x ^{3}})

What is the integral of 3^x+1/x-x^{sqrt(3)} ?
The integral of 3^x+1/x-x^{sqrt(3)} is (3^x}{ln(3)}+ln|x|-\frac{sqrt(3)-1)/2 x^{sqrt(3)+1}+C