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

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

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integral of 3xsqrt(7-3x^2)

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

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

Solution steps

\int 3 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

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

Apply u-substitution

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

= 3 \cdot \int - \frac{u}{6} d u

Take the constant out:

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

= 3 (- \frac{1}{6} \cdot \int u d u)

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = 7 - 3 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{x ( 1 + ln ( x ) )} d x

t^{2} y^{^{''}} + t y^{^{'}} + y = 0

\frac{d}{d x} (e^{4 x} + 4 e^{4 x} x)

\int \frac{2 x ^{3} + 3 x}{x ^{4} + 3 x ^{2} + 7} d x

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