What is the integral of 1/(x^2y) ?

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

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integral of 1/(x^2y)

\int \frac{1}{x ^{2} y} d y

\frac{1}{x ^{2}} ln ∣ y ∣ + C

Solution steps

\int \frac{1}{x ^{2} y} d y

Take the constant out:

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

= \frac{1}{x ^{2}} \cdot \int \frac{1}{y} d y

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= \frac{1}{x ^{2}} ln ∣ y ∣

Add a constant to the solution

= \frac{1}{x ^{2}} ln ∣ y ∣ + C

- x^{3} + 9 x^{2} - 6 x + 9

sin (2 x) cos (3 x)

\int x^{2} e^{x} cos (x) d x

\int 2 x cos (x^{2}) d x

f (x) = 0.2 x^{2} + 10, g (x) = x, x = - 7, x = 2