What is the integral of x^3 1/(sqrt(1-x^2)) ?

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

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integral of x^3 1/(sqrt(1-x^2))

\int x^{3} \frac{1}{1 - x ^{2}} d x

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

Solution steps

\int x^{3} \frac{1}{1 - x ^{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{3 - u ^{\frac{2}{3}} + 1} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{3 v ^{3}}{- v ^{2} + 1} d v

= \frac{1}{3} \cdot \int \frac{3 v ^{3}}{- v ^{2} + 1} d v

Take the constant out:

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

= \frac{1}{3} \cdot 3 \cdot \int \frac{v ^{3}}{- v ^{2} + 1} d v

Apply u-substitution

: \int w^{2} - 1 d w

= \frac{1}{3} \cdot 3 \cdot \int w^{2} - 1 d w

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}{3} \cdot 3 (\int w^{2} d w - \int 1 d w)

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

\int 1 d w = w

= \frac{1}{3} \cdot 3 (\frac{w ^{3}}{3} - w)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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