What is the integral of (x^3)/(sqrt(9x^2+16)) ?

The integral of (x^3)/(sqrt(9x^2+16)) is 1/81 (1/3 (9x^2+16)^{3/2}-16sqrt(9x^2+16))+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(9x^2+16))

\int \frac{x ^{3}}{9 x ^{2} + 16} d x

\frac{1}{81} (\frac{1}{3} (9 x^{2} + 16)^{\frac{3}{2}} - 16 9 x^{2} + 16) + C

Solution steps

\int \frac{x ^{3}}{9 x ^{2} + 16} d x

Apply u-substitution

: \int \frac{u ^{2} - 16}{81} d u

= \int \frac{u ^{2} - 16}{81} d u

Take the constant out:

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

= \frac{1}{81} \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

= \frac{1}{81} (\int u^{2} d u - \int 16 d u)

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

\int 16 d u = 16 u

= \frac{1}{81} (\frac{u ^{3}}{3} - 16 u)

Substitute back

u = 9 x^{2} + 16

= \frac{1}{81} (\frac{( 9 x ^{2} + 16 ) ^{3}}{3} - 16 9 x^{2} + 16)

Simplify

\frac{1}{81} (\frac{( 9 x ^{2} + 16 ) ^{3}}{3} - 16 9 x^{2} + 16) : \frac{1}{81} (\frac{1}{3} (9 x^{2} + 16)^{\frac{3}{2}} - 16 9 x^{2} + 16)

= \frac{1}{81} (\frac{1}{3} (9 x^{2} + 16)^{\frac{3}{2}} - 16 9 x^{2} + 16)

Add a constant to the solution

= \frac{1}{81} (\frac{1}{3} (9 x^{2} + 16)^{\frac{3}{2}} - 16 9 x^{2} + 16) + C

\int x^{6} (x^{7} + 5) d x

\int e^{3 x} d x

x^{- 1}

\int \frac{1}{x} \frac{1 - x}{1 + x} d x

\int 9 x^{2} arcsin (x) d x

What is the integral of (x^3)/(sqrt(9x^2+16)) ?
The integral of (x^3)/(sqrt(9x^2+16)) is 1/81 (1/3 (9x^2+16)^{3/2}-16sqrt(9x^2+16))+C