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

The integral of (2x^2)/(x^2+9) is 6(-arctan(x/3)+x/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

6 (- arctan (\frac{x}{3}) + \frac{x}{3}) + C

Solution steps

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

Take the constant out:

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

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

\frac{x ^{2}}{x ^{2} + 9} = - \frac{9}{x ^{2} + 9} + 1

= 2 \cdot \int - \frac{9}{x ^{2} + 9} + 1 d x

Apply Integral Substitution

= 2 \cdot \int 3 (- \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

= 2 \cdot 3 \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

= 2 \cdot 3 (- \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

= 2 \cdot 3 (- arctan (u) + u)

Substitute back

u = \frac{x}{3}

= 2 \cdot 3 (- arctan (\frac{x}{3}) + \frac{x}{3})

Simplify

= 6 (- arctan (\frac{x}{3}) + \frac{x}{3})

Add a constant to the solution

= 6 (- arctan (\frac{x}{3}) + \frac{x}{3}) + C

\int \frac{t + 5}{t ^{2} + t - 2} d t

\int (e^{4} - 5^{π}) d x

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

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

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

What is the integral of (2x^2)/(x^2+9) ?
The integral of (2x^2)/(x^2+9) is 6(-arctan(x/3)+x/3)+C