What is the integral of (x^2)/(64+x^2) ?

The integral of (x^2)/(64+x^2) is -8arctan(x/8)+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)/(64+x^2)

\int \frac{x ^{2}}{64 + x ^{2}} d x

- 8 arctan (\frac{x}{8}) + x + C

Solution steps

\int \frac{x ^{2}}{64 + x ^{2}} d x

\frac{x ^{2}}{64 + x ^{2}} = - \frac{64}{64 + x ^{2}} + 1

= \int - \frac{64}{64 + x ^{2}} + 1 d x

Apply Integral Substitution

= \int 8 (- \frac{1}{1 + u ^{2}} + 1) d u

Take the constant out:

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

= 8 \cdot \int - \frac{1}{1 + u ^{2}} + 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

= 8 (- \int \frac{1}{1 + u ^{2}} d u + \int 1 d u)

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

\int 1 d u = u

= 8 (- arctan (u) + u)

Substitute back

u = \frac{x}{8}

= 8 (- arctan (\frac{x}{8}) + \frac{x}{8})

Simplify

8 (- arctan (\frac{x}{8}) + \frac{x}{8}) : - 8 arctan (\frac{x}{8}) + x

= - 8 arctan (\frac{x}{8}) + x

Add a constant to the solution

= - 8 arctan (\frac{x}{8}) + x + C

\int (cos (θ) + sin (θ)) d θ

\int (\frac{2 x ^{2} + 7 x - 3}{x - 2}) d x

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

\int - cos (π + y) + cos (y) d y

\int (x + 2) 3 x^{2} + 4 x d x

What is the integral of (x^2)/(64+x^2) ?
The integral of (x^2)/(64+x^2) is -8arctan(x/8)+x+C