What is the integral of x(3x^2-7)^{-2} ?

The integral of x(3x^2-7)^{-2} is -1/(6(3x^2-7))+C

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integral of x(3x^2-7)^{-2}

\int x (3 x^{2} - 7)^{- 2} d x

- \frac{1}{6 ( 3 x ^{2} - 7 )} + C

Solution steps

\int x (3 x^{2} - 7)^{- 2} d x

Simplify

x (3 x^{2} - 7)^{- 2} : \frac{x}{( 3 x ^{2} - 7 ) ^{2}}

= \int \frac{x}{( 3 x ^{2} - 7 ) ^{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} - 7

= \frac{1}{6} (- \frac{1}{3 x ^{2} - 7})

Simplify

\frac{1}{6} (- \frac{1}{3 x ^{2} - 7}) : - \frac{1}{6 ( 3 x ^{2} - 7 )}

= - \frac{1}{6 ( 3 x ^{2} - 7 )}

Add a constant to the solution

= - \frac{1}{6 ( 3 x ^{2} - 7 )} + C

\frac{d}{d x} (\frac{3}{4} x^{4} - 5 x^{2} + 4 x + 1)

\frac{d}{d x} (cos^{2} (2 - x^{2}))

t^{2} y^{^{''}} - 3 t y^{^{'}} + 3 y = 0, y (1) = 5, y^{^{'}} (1) = 3

5 cos (4 x), 5 - 5 cos (4 x), [0, \frac{π}{4}]

\frac{\partial}{\partial x} (2 x - 4 y^{2} + 3 z^{2})

What is the integral of x(3x^2-7)^{-2} ?
The integral of x(3x^2-7)^{-2} is -1/(6(3x^2-7))+C