What is the integral of (6x)/(sqrt(7-3x^2)) ?

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

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integral of (6x)/(sqrt(7-3x^2))

\int \frac{6 x}{7 - 3 x ^{2}} d x

- 2 7 - 3 x^{2} + C

Solution steps

\int \frac{6 x}{7 - 3 x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{7 - 3 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{6 u} d u

= 6 \cdot \int - \frac{1}{6 u} d u

Take the constant out:

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

= 6 (- \frac{1}{6} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 6 (- \frac{1}{6} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 7 - 3 x^{2}

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

Simplify

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

= - 2 7 - 3 x^{2}

Add a constant to the solution

= - 2 7 - 3 x^{2} + C

\int_{0}^{2} (4 - x^{2})^{2} d x

x \to 1 - lim (4)

y = \frac{1}{x - 5}, at x = 10

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

x \to \infty lim (1 - \frac{4}{x})

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