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

The integral of (x^2)/(x^2+16) is -4arctan(x/4)+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(x^2+16)

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

- 4 arctan (\frac{x}{4}) + x + C

Solution steps

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

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

= \int - \frac{16}{x ^{2} + 16} + 1 d x

Apply Integral Substitution

= \int 4 (- \frac{1}{u ^{2} + 1} + 1) d u

Take the constant out:

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

= 4 \cdot \int - \frac{1}{u ^{2} + 1} + 1 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 \frac{1}{u ^{2} + 1} d u + \int 1 d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= 4 (- arctan (u) + u)

Substitute back

u = \frac{x}{4}

= 4 (- arctan (\frac{x}{4}) + \frac{x}{4})

Simplify

4 (- arctan (\frac{x}{4}) + \frac{x}{4}) : - 4 arctan (\frac{x}{4}) + x

= - 4 arctan (\frac{x}{4}) + x

Add a constant to the solution

= - 4 arctan (\frac{x}{4}) + x + C

n = 1 \sum \infty \frac{1000}{5 ^{n}}

5 + 4 e^{- 2 t}

\int (2 x) d x

\int x + y d x

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

What is the integral of (x^2)/(x^2+16) ?
The integral of (x^2)/(x^2+16) is -4arctan(x/4)+x+C