What is the integral of x/((3x^2-4)^2) ?

The integral of x/((3x^2-4)^2) is -1/(6(3x^2-4))+C

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integral of x/((3x^2-4)^2)

\int \frac{x}{( 3 x ^{2} - 4 ) ^{2}} d x

- \frac{1}{6 ( 3 x ^{2} - 4 )} + C

Solution steps

\int \frac{x}{( 3 x ^{2} - 4 ) ^{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} - 4

= \frac{1}{6} (- \frac{1}{3 x ^{2} - 4})

Simplify

\frac{1}{6} (- \frac{1}{3 x ^{2} - 4}) : - \frac{1}{6 ( 3 x ^{2} - 4 )}

= - \frac{1}{6 ( 3 x ^{2} - 4 )}

Add a constant to the solution

= - \frac{1}{6 ( 3 x ^{2} - 4 )} + C

4 x^{2}, 16

f (x) = 7 x^{4} + 5

\frac{\partial}{\partial y} (4 - x^{2} - 9 y^{2})

\int 4 x y e^{- (2 x + y)} d y

y^{^{'}} = 6 x - y

What is the integral of x/((3x^2-4)^2) ?
The integral of x/((3x^2-4)^2) is -1/(6(3x^2-4))+C