What is the integral of (x^3)/((49-4x^2)^{3/2)} ?

The integral of (x^3)/((49-4x^2)^{3/2)} is (-2x^2+49)/(8(-4x^2+49)^{1/2)}+C

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integral of (x^3)/((49-4x^2)^{3/2)}

\int \frac{x ^{3}}{( 49 - 4 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{- 2 x ^{2} + 49}{8 ( - 4 x ^{2} + 49 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x ^{3}}{( 49 - 4 x ^{2} ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{u - 49}{32 u ^{\frac{3}{2}}} d u

= \int \frac{u - 49}{32 u ^{\frac{3}{2}}} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int \frac{u - 49}{u ^{\frac{3}{2}}} d u

Expand

\frac{u - 49}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{49}{u ^{\frac{3}{2}}}

= \frac{1}{32} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{49}{u ^{\frac{3}{2}}} 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

= \frac{1}{32} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{49}{u ^{\frac{3}{2}}} d u)

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

\int \frac{49}{u ^{\frac{3}{2}}} d u = - \frac{98}{u ^{\frac{1}{2}}}

= \frac{1}{32} (2 u^{\frac{1}{2}} - (- \frac{98}{u ^{\frac{1}{2}}}))

Substitute back

u = 49 - 4 x^{2}

= \frac{1}{32} (2 (49 - 4 x^{2})^{\frac{1}{2}} - (- \frac{98}{( 49 - 4 x ^{2} ) ^{\frac{1}{2}}}))

Simplify

\frac{1}{32} (2 (49 - 4 x^{2})^{\frac{1}{2}} - (- \frac{98}{( 49 - 4 x ^{2} ) ^{\frac{1}{2}}})) : \frac{- 2 x ^{2} + 49}{8 ( - 4 x ^{2} + 49 ) ^{\frac{1}{2}}}

= \frac{- 2 x ^{2} + 49}{8 ( - 4 x ^{2} + 49 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{- 2 x ^{2} + 49}{8 ( - 4 x ^{2} + 49 ) ^{\frac{1}{2}}} + C

y^{^{'}} = \frac{( 5 x ^{2} + y ^{2} )}{x y}

\int_{0}^{\frac{1}{2} - x} 384 x y d y

\int x (4 x^{2} + 7)^{9} d x

\frac{d y}{d x} = \frac{y}{1 + x ^{2}}

What is the integral of (x^3)/((49-4x^2)^{3/2)} ?
The integral of (x^3)/((49-4x^2)^{3/2)} is (-2x^2+49)/(8(-4x^2+49)^{1/2)}+C