What is the integral of (9x^2)/((7-3x^3)^{1/2)} ?

The integral of (9x^2)/((7-3x^3)^{1/2)} is (3x^3)/((7-3x^3)^{1/2)}-(-3x^3+14)/((-3x^3+7)^{1/2)}+C

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integral of (9x^2)/((7-3x^3)^{1/2)}

\int \frac{9 x ^{2}}{( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} d x

\frac{3 x ^{3}}{( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} - \frac{- 3 x ^{3} + 14}{( - 3 x ^{3} + 7 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{9 x ^{2}}{( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} d x

Take the constant out:

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

= 9 \cdot \int \frac{x ^{2}}{( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} d x

Apply Integration By Parts

= 9 (\frac{x ^{3}}{3 ( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} - \int \frac{3 x ^{5}}{2 ( - 3 x ^{3} + 7 ) ^{\frac{3}{2}}} d x)

\int \frac{3 x ^{5}}{2 ( - 3 x ^{3} + 7 ) ^{\frac{3}{2}}} d x = \frac{- 3 x ^{3} + 14}{9 ( - 3 x ^{3} + 7 ) ^{\frac{1}{2}}}

= 9 (\frac{x ^{3}}{3 ( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} - \frac{- 3 x ^{3} + 14}{9 ( - 3 x ^{3} + 7 ) ^{\frac{1}{2}}})

Simplify

9 (\frac{x ^{3}}{3 ( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} - \frac{- 3 x ^{3} + 14}{9 ( - 3 x ^{3} + 7 ) ^{\frac{1}{2}}}) : \frac{3 x ^{3}}{( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} - \frac{- 3 x ^{3} + 14}{( - 3 x ^{3} + 7 ) ^{\frac{1}{2}}}

= \frac{3 x ^{3}}{( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} - \frac{- 3 x ^{3} + 14}{( - 3 x ^{3} + 7 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{3 x ^{3}}{( 7 - 3 x ^{3} ) ^{\frac{1}{2}}} - \frac{- 3 x ^{3} + 14}{( - 3 x ^{3} + 7 ) ^{\frac{1}{2}}} + C

What is the integral of (9x^2)/((7-3x^3)^{1/2)} ?
The integral of (9x^2)/((7-3x^3)^{1/2)} is (3x^3)/((7-3x^3)^{1/2)}-(-3x^3+14)/((-3x^3+7)^{1/2)}+C