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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

x \to 8 lim (\frac{x}{x - 8})

\frac{d y}{d x} y = (x^{2} + 1)^{3}

y = 3 + 5 x^{2} - 2 x^{3}

\frac{\partial}{\partial x} ((x - 2)^{2} + (5 - x - y)^{2})

x \to 7 lim (- x^{2} + 3 x - 8)