What is the integral of 1/((a+bx)^2) ?

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

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integral of 1/((a+bx)^2)

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = a + b x

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

Simplify

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

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

Add a constant to the solution

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

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