What is the integral of x/((3x^2+8)^2) ?

The integral of x/((3x^2+8)^2) is -1/(6(3x^2+8))+C

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integral of x/((3x^2+8)^2)

\int \frac{x}{( 3 x ^{2} + 8 ) ^{2}} d x

- \frac{1}{6 ( 3 x ^{2} + 8 )} + C

Solution steps

\int \frac{x}{( 3 x ^{2} + 8 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{6 u ^{2}} d u

= \int \frac{1}{6 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{6} (- \frac{1}{u})

Substitute back

u = 3 x^{2} + 8

= \frac{1}{6} (- \frac{1}{3 x ^{2} + 8})

Simplify

\frac{1}{6} (- \frac{1}{3 x ^{2} + 8}) : - \frac{1}{6 ( 3 x ^{2} + 8 )}

= - \frac{1}{6 ( 3 x ^{2} + 8 )}

Add a constant to the solution

= - \frac{1}{6 ( 3 x ^{2} + 8 )} + C

f (x) = ln (x + 3)

\int (3 x + 1) d x

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

\frac{s}{s + 1000}

\frac{d}{d y} (\frac{e ^{y} + e ^{- y}}{2})

What is the integral of x/((3x^2+8)^2) ?
The integral of x/((3x^2+8)^2) is -1/(6(3x^2+8))+C