What is the integral of 1/x-x^2+sin(x)-3^x ?

The integral of 1/x-x^2+sin(x)-3^x is ln|x|-(x^3)/3-cos(x)-(3^x)/(ln(3))+C

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integral of 1/x-x^2+sin(x)-3^x

\int \frac{1}{x} - x^{2} + sin (x) - 3^{x} d x

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

Solution steps

\int \frac{1}{x} - x^{2} + sin (x) - 3^{x} 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 \frac{1}{x} d x - \int x^{2} d x + \int sin (x) d x - \int 3^{x} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int sin (x) d x = - cos (x)

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

= ln ∣ x ∣ - \frac{x ^{3}}{3} - cos (x) - \frac{3 ^{x}}{ln ( 3 )}

Add a constant to the solution

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

\frac{d y}{d x} + 2 y = 9 e^{x}

\frac{\partial}{\partial y} (3 x + 8 y + 2)

\frac{\partial}{\partial x} (y ln (y))

\int sin (8 x) cos (6 x) d x

2 x y^{^{'}} - y = x^{2}

What is the integral of 1/x-x^2+sin(x)-3^x ?
The integral of 1/x-x^2+sin(x)-3^x is ln|x|-(x^3)/3-cos(x)-(3^x)/(ln(3))+C