What is the integral of (4x^3)/(sqrt(x^2-16)) ?

The integral of (4x^3)/(sqrt(x^2-16)) is 4(1/3 (x^2-16)^{3/2}+16sqrt(x^2-16))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (4x^3)/(sqrt(x^2-16))

\int \frac{4 x ^{3}}{x ^{2} - 16} d x

4 (\frac{1}{3} (x^{2} - 16)^{\frac{3}{2}} + 16 x^{2} - 16) + C

Solution steps

\int \frac{4 x ^{3}}{x ^{2} - 16} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{3}}{x ^{2} - 16} d x

Apply u-substitution

= 4 \cdot \int u^{2} + 16 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

= 4 (\int u^{2} d u + \int 16 d u)

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

\int 16 d u = 16 u

= 4 (\frac{u ^{3}}{3} + 16 u)

Substitute back

u = x^{2} - 16

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

Simplify

4 (\frac{( x ^{2} - 16 ) ^{3}}{3} + 16 x^{2} - 16) : 4 (\frac{1}{3} (x^{2} - 16)^{\frac{3}{2}} + 16 x^{2} - 16)

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

Add a constant to the solution

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

What is the integral of (4x^3)/(sqrt(x^2-16)) ?
The integral of (4x^3)/(sqrt(x^2-16)) is 4(1/3 (x^2-16)^{3/2}+16sqrt(x^2-16))+C