What is the integral of 1/((3x+7)^2) ?

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

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integral of 1/((3x+7)^2)

\int \frac{1}{( 3 x + 7 ) ^{2}} d x

- \frac{1}{3 ( 3 x + 7 )} + C

Solution steps

\int \frac{1}{( 3 x + 7 ) ^{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 + 7

= \frac{1}{3} (- \frac{1}{3 x + 7})

Simplify

\frac{1}{3} (- \frac{1}{3 x + 7}) : - \frac{1}{3 ( 3 x + 7 )}

= - \frac{1}{3 ( 3 x + 7 )}

Add a constant to the solution

= - \frac{1}{3 ( 3 x + 7 )} + C

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

e^{- 13 x}

\frac{d y}{d x} = \frac{( x ^{2} - 3 y ^{2} )}{2 x y}

\frac{\partial}{\partial x} (\frac{4 s + t}{3 s - 5 t})

y^{^{''}} + 49 y = 0, y (0) = 4, y^{^{''}} (0) = 6

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