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

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

\int \frac{2 x - 3}{x ^{2} + 9} d x

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

Solution steps

\int \frac{2 x - 3}{x ^{2} + 9} d x

Apply Integral Substitution

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

Expand

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

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

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

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

= ln u^{2} + 1 - arctan (u)

Substitute back

u = \frac{x}{3}

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

(\frac{x}{3})^{2} = \frac{x ^{2}}{9}

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

Add a constant to the solution

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

\int (2 x^{2} + 6)^{2} d x

\int x 4 x^{5} d x

\int \frac{1}{5} x e^{- \frac{x}{5}} d x

\int - sin (x) e^{c o s (x)} d x

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

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