What is the integral of 1/((4+x)^2) ?

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

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integral of 1/((4+x)^2)

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

- \frac{1}{4 + x} + C

Solution steps

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

= - \frac{1}{4 + x}

Add a constant to the solution

= - \frac{1}{4 + x} + C

y = - x^{2} + 12 x - 32, y = 0

x \to 4 lim (\frac{15}{x - 4})

(\frac{d y}{d x}) - y = 4 e^{x} + 24 e^{(4 x)}

\frac{d}{d x} (- \frac{1}{2} cot^{3} (x))

\int - e^{y} d y