What is the integral of (2x+1)^{-2} ?

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

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integral of (2x+1)^{-2}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

\int 6 e^{- 2 x} d x

\frac{d}{d x} ((1 - x)^{2} + 10 (y - x^{2})^{2})

\frac{d}{d x} (3 x - sin^{2} (4 x))

x \to 0 lim (\frac{x}{e ^{7 x} - 1})

y^{^{'}} + 4 y = 32

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