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

The integral of ((2+x))/(9+x^2) is 2/3 arctan(x/3)+1/2 ln|(x^2)/9+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((2+x))/(9+x^2)

\int \frac{( 2 + x )}{9 + x ^{2}} d x

\frac{2}{3} arctan (\frac{x}{3}) + \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + C

Solution steps

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

Apply Integral Substitution

= \int \frac{2 + 3 u}{3 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{2 + 3 u}{u ^{2} + 1} d u

Expand

\frac{2 + 3 u}{u ^{2} + 1} : \frac{2}{u ^{2} + 1} + \frac{3 u}{u ^{2} + 1}

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

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

\int \frac{3 u}{u ^{2} + 1} d u = \frac{3}{2} ln u^{2} + 1

= \frac{1}{3} (2 arctan (u) + \frac{3}{2} ln u^{2} + 1)

Substitute back

u = \frac{x}{3}

= \frac{1}{3} (2 arctan (\frac{x}{3}) + \frac{3}{2} ln (\frac{x}{3})^{2} + 1)

Simplify

\frac{1}{3} (2 arctan (\frac{x}{3}) + \frac{3}{2} ln (\frac{x}{3})^{2} + 1) : \frac{2}{3} arctan (\frac{x}{3}) + \frac{1}{2} ln \frac{x ^{2}}{9} + 1

= \frac{2}{3} arctan (\frac{x}{3}) + \frac{1}{2} ln \frac{x ^{2}}{9} + 1

Add a constant to the solution

= \frac{2}{3} arctan (\frac{x}{3}) + \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + C

\int_{0}^{1} x (\frac{2 ( x + 3 )}{7}) d x

f (x) = - 3 cos (x)

\int \frac{x ^{2}}{4 x ^{3} + 5} d x

f (t) = 1 + cos (t)

\int_{- \infty}^{8} r e^{\frac{r}{4}} d r

What is the integral of ((2+x))/(9+x^2) ?
The integral of ((2+x))/(9+x^2) is 2/3 arctan(x/3)+1/2 ln|(x^2)/9+1|+C