What is the integral of (x^2)/((15+4x-4x^2)^{3/2)} ?

The integral of (x^2)/((15+4x-4x^2)^{3/2)} is

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integral of (x^2)/((15+4x-4x^2)^{3/2)}

\int \frac{x ^{2}}{( 15 + 4 x - 4 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{1}{32} (4 (- arcsin (\frac{1}{2} (x - \frac{1}{2})) + \frac{2 x - 1}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}}) + \frac{8}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}} + \frac{2 x - 1}{4 ( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{x ^{2}}{( 15 + 4 x - 4 x ^{2} ) ^{\frac{3}{2}}} d x

Complete the square

15 + 4 x - 4 x^{2} : - 4 (x - \frac{1}{2})^{2} + 16

= \int \frac{x ^{2}}{( - 4 ( x - \frac{1}{2} ) ^{2} + 16 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{( 2 u + 1 ) ^{2}}{32 ( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u

= \int \frac{( 2 u + 1 ) ^{2}}{32 ( - u ^{2} + 4 ) ^{\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{( 2 u + 1 ) ^{2}}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u

Expand

\frac{( 2 u + 1 ) ^{2}}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} : \frac{4 u ^{2}}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{4 u}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{1}{( - u ^{2} + 4 ) ^{\frac{3}{2}}}

= \frac{1}{32} \cdot \int \frac{4 u ^{2}}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{4 u}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{1}{( - u ^{2} + 4 ) ^{\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{4 u ^{2}}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u + \int \frac{4 u}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u + \int \frac{1}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u)

\int \frac{4 u ^{2}}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u = 4 (- arcsin (\frac{1}{2} u) + \frac{u}{( 4 - u ^{2} ) ^{\frac{1}{2}}})

\int \frac{4 u}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u = \frac{4}{( - u ^{2} + 4 ) ^{\frac{1}{2}}}

\int \frac{1}{( - u ^{2} + 4 ) ^{\frac{3}{2}}} d u = \frac{u}{4 ( 4 - u ^{2} ) ^{\frac{1}{2}}}

= \frac{1}{32} (4 (- arcsin (\frac{1}{2} u) + \frac{u}{( 4 - u ^{2} ) ^{\frac{1}{2}}}) + \frac{4}{( - u ^{2} + 4 ) ^{\frac{1}{2}}} + \frac{u}{4 ( 4 - u ^{2} ) ^{\frac{1}{2}}})

Substitute back

u = x - \frac{1}{2}

= \frac{1}{32} 4 - arcsin (\frac{1}{2} (x - \frac{1}{2})) + \frac{x - \frac{1}{2}}{( 4 - ( x - \frac{1}{2} ) ^{2} ) ^{\frac{1}{2}}} + \frac{4}{( - ( x - \frac{1}{2} ) ^{2} + 4 ) ^{\frac{1}{2}}} + \frac{x - \frac{1}{2}}{4 ( 4 - ( x - \frac{1}{2} ) ^{2} ) ^{\frac{1}{2}}}

Simplify

\frac{1}{32} 4 - arcsin (\frac{1}{2} (x - \frac{1}{2})) + \frac{x - \frac{1}{2}}{( 4 - ( x - \frac{1}{2} ) ^{2} ) ^{\frac{1}{2}}} + \frac{4}{( - ( x - \frac{1}{2} ) ^{2} + 4 ) ^{\frac{1}{2}}} + \frac{x - \frac{1}{2}}{4 ( 4 - ( x - \frac{1}{2} ) ^{2} ) ^{\frac{1}{2}}} : \frac{1}{32} (4 (- arcsin (\frac{1}{2} (x - \frac{1}{2})) + \frac{2 x - 1}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}}) + \frac{8}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}} + \frac{2 x - 1}{4 ( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}})

= \frac{1}{32} (4 (- arcsin (\frac{1}{2} (x - \frac{1}{2})) + \frac{2 x - 1}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}}) + \frac{8}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}} + \frac{2 x - 1}{4 ( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{32} (4 (- arcsin (\frac{1}{2} (x - \frac{1}{2})) + \frac{2 x - 1}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}}) + \frac{8}{( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}} + \frac{2 x - 1}{4 ( - 4 x ^{2} + 4 x + 15 ) ^{\frac{1}{2}}}) + C

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What is the integral of (x^2)/((15+4x-4x^2)^{3/2)} ?
The integral of (x^2)/((15+4x-4x^2)^{3/2)} is