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

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

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integral of 1/((x+2)(x+2))

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

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

Solution steps

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

Take the partial fraction of

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 2

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

Add a constant to the solution

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

f (x) = sin^{5} (x)

\frac{d}{d t} (arcsin (2 t))

\frac{d}{d x} (\frac{x ^{4}}{5})

f (x) = - x^{3} - 3 x^{2} - 3, at x = - 3

x \to - 1 lim (\frac{x + 2}{x ^{2} + x})

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