What is the integral of (3x^3)/(sqrt(49-x^2)) ?

The integral of (3x^3)/(sqrt(49-x^2)) is (49-x^2)^{3/2}-147sqrt(49-x^2)+C

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integral of (3x^3)/(sqrt(49-x^2))

\int \frac{3 x ^{3}}{49 - x ^{2}} d x

(49 - x^{2})^{\frac{3}{2}} - 147 49 - x^{2} + C

Solution steps

\int \frac{3 x ^{3}}{49 - 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 \frac{x ^{3}}{49 - x ^{2}} d x

Apply u-substitution

: \int u^{2} - 49 d u

= 3 \cdot \int u^{2} - 49 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int u^{2} d u - \int 49 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 49 d u = 49 u

= 3 (\frac{u ^{3}}{3} - 49 u)

Substitute back

u = 49 - x^{2}

= 3 (\frac{( 49 - x ^{2} ) ^{3}}{3} - 49 49 - x^{2})

Simplify

3 (\frac{( 49 - x ^{2} ) ^{3}}{3} - 49 49 - x^{2}) : (49 - x^{2})^{\frac{3}{2}} - 147 49 - x^{2}

= (49 - x^{2})^{\frac{3}{2}} - 147 49 - x^{2}

Add a constant to the solution

= (49 - x^{2})^{\frac{3}{2}} - 147 49 - x^{2} + C

x^{2} - 3 x - 2

\frac{d}{d x} (arccos (\frac{x + 1}{3}))

y^{^{'}} = x + y, y (0) = 0

\int u^{2} d u

\int_{0}^{3} 3 x d x

What is the integral of (3x^3)/(sqrt(49-x^2)) ?
The integral of (3x^3)/(sqrt(49-x^2)) is (49-x^2)^{3/2}-147sqrt(49-x^2)+C