What is the integral of 1/((ax+b)^2) ?

The integral of 1/((ax+b)^2) is -1/(a(ax+b))+C

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integral of 1/((ax+b)^2)

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

- \frac{1}{a ( a x + b )} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = a x + b

= \frac{1}{a} (- \frac{1}{a x + b})

Simplify

\frac{1}{a} (- \frac{1}{a x + b}) : - \frac{1}{a ( a x + b )}

= - \frac{1}{a ( a x + b )}

Add a constant to the solution

= - \frac{1}{a ( a x + b )} + C

x \to 3 lim (\frac{1}{∣ x - 3 ∣})

\frac{d}{d x} (x^{2} \cdot 2 x)

\frac{d}{d x} (x^{2} ln (\frac{x}{2}))

x \to \frac{2}{3} lim (\frac{3 x - 2}{3 x ^{2} - 11 x + 6})

\frac{\partial}{\partial z} (x y + yz + z x)

What is the integral of 1/((ax+b)^2) ?
The integral of 1/((ax+b)^2) is -1/(a(ax+b))+C