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

The integral of 1/(x^2yz) is -1/(yzx)+C

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integral of 1/(x^2yz)

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

- \frac{1}{yz x} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{x}

= \frac{1}{yz} (- \frac{1}{x})

Simplify

\frac{1}{yz} (- \frac{1}{x}) : - \frac{1}{yz x}

= - \frac{1}{yz x}

Add a constant to the solution

= - \frac{1}{yz x} + C

ln (4 - 3 x)

x y^{2} y^{^{'}} = x + 3

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

y^{^{'}} = \frac{1}{sec ^{2} ( y )}

\frac{d}{d x} (\frac{sin ( 3 x )}{cos ( 3 x )})