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

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

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

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

- \frac{1}{3 ( 3 x + 2 )} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u ^{2}} d u

= \int \frac{1}{3 u ^{2}} d u

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{3} (- \frac{1}{u})

Substitute back

u = 3 x + 2

= \frac{1}{3} (- \frac{1}{3 x + 2})

Simplify

\frac{1}{3} (- \frac{1}{3 x + 2}) : - \frac{1}{3 ( 3 x + 2 )}

= - \frac{1}{3 ( 3 x + 2 )}

Add a constant to the solution

= - \frac{1}{3 ( 3 x + 2 )} + C

n = 2 \sum \infty \frac{n + 4}{n !}

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

2 y = 5, y = 3, 2 y + 3 x = 8

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

\frac{\partial}{\partial x} (x y^{2} arctan (z))

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