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

The integral of (x^3)/(16+x^2) is 1/2 (16+x^2-16ln|16+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 (x^3)/(16+x^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{u - 16}{u} : 1 - \frac{16}{u}

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

\int 1 d u = u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{1}{2} (u - 16 ln ∣ u ∣)

Substitute back

u = 16 + x^{2}

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

Add a constant to the solution

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

\int (1 - e^{x})^{2} d x

\int x^{2} 7 + x^{3} d x

\int sin^{6} (\frac{x}{2}) cos (\frac{x}{2}) d x

\int tan (x) (ln (cos (x))) d x

\int x^{3} (5 + x^{4})^{4} d x

What is the integral of (x^3)/(16+x^2) ?
The integral of (x^3)/(16+x^2) is 1/2 (16+x^2-16ln|16+x^2|)+C