What is the integral of 1/(x^3y^2) ?

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

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integral of 1/(x^3y^2)

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

- \frac{1}{2 y ^{2} x ^{2}} + C

Solution steps

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

Take the constant out:

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

= \frac{1}{y ^{2}} \cdot \int \frac{1}{x ^{3}} d x

Apply the Power Rule

: - \frac{1}{2 x ^{2}}

= \frac{1}{y ^{2}} (- \frac{1}{2 x ^{2}})

Simplify

\frac{1}{y ^{2}} (- \frac{1}{2 x ^{2}}) : - \frac{1}{2 y ^{2} x ^{2}}

= - \frac{1}{2 y ^{2} x ^{2}}

Add a constant to the solution

= - \frac{1}{2 y ^{2} x ^{2}} + C

\int (x^{3} + 2)^{2} \cdot 3 x^{2} d x

\int 2 3 x d x

\int - cos (x) + 2 d x

\int \frac{1}{4 + cos ( x )} d x

\int ln x^{2} + 1 d x

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