What is the integral of (4x+5)^{-2} ?

The integral of (4x+5)^{-2} is -1/(4(4x+5))+C

integral of (4x+5)^{-2}

\int (4 x + 5)^{- 2} d x

- \frac{1}{4 ( 4 x + 5 )} + C

Solution steps

\int (4 x + 5)^{- 2} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 4 x + 5 ) ^{2}} d x

Apply u-substitution

= \int \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= \frac{1}{4} (- \frac{1}{u})

Substitute back

u = 4 x + 5

= \frac{1}{4} (- \frac{1}{4 x + 5})

Simplify

\frac{1}{4} (- \frac{1}{4 x + 5}) : - \frac{1}{4 ( 4 x + 5 )}

= - \frac{1}{4 ( 4 x + 5 )}

Add a constant to the solution

= - \frac{1}{4 ( 4 x + 5 )} + C

\int cos (3 x) x d x

p a r i t y y = x^{x^{x}}

\int_{0}^{t} 1 d t

d er i v a t i v e \frac{1}{4 π ( 3 + 2 ) ^{2}} (- 11)

x \to 1 - lim (1)

What is the integral of (4x+5)^{-2} ?
The integral of (4x+5)^{-2} is -1/(4(4x+5))+C