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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^3)/(sqrt(x^2+4))

\int \frac{4 x ^{3}}{x ^{2} + 4} d x

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

Solution steps

\int \frac{4 x ^{3}}{x ^{2} + 4} 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} + 4} d x

Apply u-substitution

: \int u^{2} - 4 d u

= 4 \cdot \int u^{2} - 4 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 4 d u)

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

\int 4 d u = 4 u

= 4 (\frac{u ^{3}}{3} - 4 u)

Substitute back

u = x^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( arctan ( 5 x ) )}{1 + 25 x ^{2}} d x

\int (12 x^{2} e^{- 4 x y} - 5 x + 1) d y

\int x^{2} + 36 d x

\int \frac{x}{( ln ( x ) ) ^{2}} d x

\int (x + e^{x}) d x

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