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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

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

\int 1 d u = u

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

f (x) = - x + \frac{6}{x} + 1

\frac{s - 1}{s ^{2} - 2 s + 5}

\int \frac{4}{t} d t

2 y^{^{''}} + 32 y = 0

y = x^{3} - 6 x^{2} + 7 x + 3

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