What is the integral of 1/((x+4)^3) ?

The integral of 1/((x+4)^3) is -1/(2(x+4)^2)+C

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integral of 1/((x+4)^3)

\int \frac{1}{( x + 4 ) ^{3}} d x

- \frac{1}{2 ( x + 4 ) ^{2}} + C

Solution steps

\int \frac{1}{( x + 4 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= - \frac{1}{2 u ^{2}}

Substitute back

u = x + 4

= - \frac{1}{2 ( x + 4 ) ^{2}}

Add a constant to the solution

= - \frac{1}{2 ( x + 4 ) ^{2}} + C

y^{^{''}} - 6 y^{^{'}} + 7 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\int \frac{1}{x ( 3 x + 1 )} d x

\frac{d}{d t} (sin^{4} (t))

(t^{2} - 20 t + 75) \frac{d y}{d t} = y

f (x) = 6 x^{2} e^{6 x} + 2 x e^{6 x}

What is the integral of 1/((x+4)^3) ?
The integral of 1/((x+4)^3) is -1/(2(x+4)^2)+C