What is the integral of (x^3)/(sqrt(4x^2+9)) ?

The integral of (x^3)/(sqrt(4x^2+9)) is 1/24 (2x^2-9)sqrt(4x^2+9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(sqrt(4x^2+9))

\int \frac{x ^{3}}{4 x ^{2} + 9} d x

\frac{1}{24} (2 x^{2} - 9) 4 x^{2} + 9 + C

Solution steps

\int \frac{x ^{3}}{4 x ^{2} + 9} d x

Apply u-substitution

: \int \frac{u ^{2} - 9}{16} d u

= \int \frac{u ^{2} - 9}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u^{2} - 9 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

= \frac{1}{16} (\int u^{2} d u - \int 9 d u)

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

\int 9 d u = 9 u

= \frac{1}{16} (\frac{u ^{3}}{3} - 9 u)

Substitute back

u = 4 x^{2} + 9

= \frac{1}{16} (\frac{( 4 x ^{2} + 9 ) ^{3}}{3} - 9 4 x^{2} + 9)

Simplify

\frac{1}{16} (\frac{( 4 x ^{2} + 9 ) ^{3}}{3} - 9 4 x^{2} + 9) : \frac{1}{24} (2 x^{2} - 9) 4 x^{2} + 9

= \frac{1}{24} (2 x^{2} - 9) 4 x^{2} + 9

Add a constant to the solution

= \frac{1}{24} (2 x^{2} - 9) 4 x^{2} + 9 + C

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

\int \frac{1}{( x ^{2} + a ) \frac{3}{2}} d x

\int 3^{x} cos (2 x) d x

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

\int \frac{sin ( θ ) cos ( θ )}{3 + cos ^{2} ( θ )} d θ

What is the integral of (x^3)/(sqrt(4x^2+9)) ?
The integral of (x^3)/(sqrt(4x^2+9)) is 1/24 (2x^2-9)sqrt(4x^2+9)+C