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

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

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

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

- \frac{1}{2 ( x + 3 ) ^{2}} + C

Solution steps

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

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

Add a constant to the solution

= - \frac{1}{2 ( x + 3 ) ^{2}} + C

\int_{- 1}^{2} (y + 2) - (y^{2}) d y

\int tan (x) x d x

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 7 )} d x

\int \frac{8}{θ ^{2}} cos (\frac{8}{θ}) d θ

y = ln (8 - 4 x)

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