What is the integral of 7x^2y+3/x-1 ?

The integral of 7x^2y+3/x-1 is -x+3ln|x|+(7yx^3)/3+C

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integral of 7x^2y+3/x-1

\int 7 x^{2} y + \frac{3}{x} - 1 d x

- x + 3 ln ∣ x ∣ + \frac{7 y x ^{3}}{3} + C

Solution steps

\int 7 x^{2} y + \frac{3}{x} - 1 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 1 d x + \int \frac{3}{x} d x + \int 7 y x^{2} d x

\int 1 d x = x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int 7 y x^{2} d x = \frac{7 y x ^{3}}{3}

= - x + 3 ln ∣ x ∣ + \frac{7 y x ^{3}}{3}

Add a constant to the solution

= - x + 3 ln ∣ x ∣ + \frac{7 y x ^{3}}{3} + C

\frac{\partial}{\partial x} (12 y^{2} x^{2} - 6 y x)

\frac{d}{d x} (sin^{4} (4 x))

\frac{d x}{d y} = 3 x (5 - x)

(3 t + 6 y + 3) = \frac{d y}{d t}

x

What is the integral of 7x^2y+3/x-1 ?
The integral of 7x^2y+3/x-1 is -x+3ln|x|+(7yx^3)/3+C