What is the integral of (2x+1)/(x^2+2x+82) ?

The integral of (2x+1)/(x^2+2x+82) is ln|x^2+2x+82|-1/9 arctan(1/9 (x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+1)/(x^2+2x+82)

\int \frac{2 x + 1}{x ^{2} + 2 x + 82} d x

ln x^{2} + 2 x + 82 - \frac{1}{9} arctan (\frac{1}{9} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 82} d x

Complete the square

x^{2} + 2 x + 82 : (x + 1)^{2} + 81

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 81} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 81} d u

= \int \frac{2 u - 1}{u ^{2} + 81} d u

Expand

\frac{2 u - 1}{u ^{2} + 81} : \frac{2 u}{u ^{2} + 81} - \frac{1}{u ^{2} + 81}

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} + 81} d u - \int \frac{1}{u ^{2} + 81} d u

\int \frac{2 u}{u ^{2} + 81} d u = ln u^{2} + 81

\int \frac{1}{u ^{2} + 81} d u = \frac{1}{9} arctan (\frac{u}{9})

= ln u^{2} + 81 - \frac{1}{9} arctan (\frac{u}{9})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 81 - \frac{1}{9} arctan (\frac{x + 1}{9})

Simplify

ln (x + 1)^{2} + 81 - \frac{1}{9} arctan (\frac{x + 1}{9}) : ln x^{2} + 2 x + 82 - \frac{1}{9} arctan (\frac{1}{9} (x + 1))

= ln x^{2} + 2 x + 82 - \frac{1}{9} arctan (\frac{1}{9} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 82 - \frac{1}{9} arctan (\frac{1}{9} (x + 1)) + C

x \to 2 lim (4 x^{2})

\frac{d}{d x} (e^{x^{2} - 4} + 4 x + 11)

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

\int (ln (7 x))^{2} d x

\int_{0}^{x^{2}} e^{t^{2}} d t

What is the integral of (2x+1)/(x^2+2x+82) ?
The integral of (2x+1)/(x^2+2x+82) is ln|x^2+2x+82|-1/9 arctan(1/9 (x+1))+C