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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

\int \frac{2 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

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

Apply u-substitution

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

= 2 \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

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

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

\int 1 d u = u

= 2 (\frac{u ^{3}}{3} - u)

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int (4 x - 6) x^{2} - 3 x d x

\int e^{\frac{t}{100}} d t

\int 6 x^{- \frac{2}{3}} - 2 x^{\frac{5}{3}} d x

\int \frac{1}{lo g _{e} ( x )} d x

\int (5 x^{4} + 6 x + 9) d x

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