What is the integral of (3x+5)/(x^2+2x+10) ?

The integral of (3x+5)/(x^2+2x+10) is 3/2 ln|x^2+2x+10|+2/3 arctan(1/3 (x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x+5)/(x^2+2x+10)

\int \frac{3 x + 5}{x ^{2} + 2 x + 10} d x

\frac{3}{2} ln x^{2} + 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x + 1)) + C

Solution steps

\int \frac{3 x + 5}{x ^{2} + 2 x + 10} d x

Complete the square

x^{2} + 2 x + 10 : (x + 1)^{2} + 9

= \int \frac{3 x + 5}{( x + 1 ) ^{2} + 9} d x

Apply u-substitution

= \int \frac{3 u + 2}{u ^{2} + 9} d u

Expand

\frac{3 u + 2}{u ^{2} + 9} : \frac{3 u}{u ^{2} + 9} + \frac{2}{u ^{2} + 9}

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{3 u}{u ^{2} + 9} d u + \int \frac{2}{u ^{2} + 9} d u

\int \frac{3 u}{u ^{2} + 9} d u = \frac{3}{2} ln u^{2} + 9

\int \frac{2}{u ^{2} + 9} d u = \frac{2}{3} arctan (\frac{u}{3})

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

Substitute back

u = x + 1

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

Simplify

\frac{3}{2} ln (x + 1)^{2} + 9 + \frac{2}{3} arctan (\frac{x + 1}{3}) : \frac{3}{2} ln x^{2} + 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x + 1))

= \frac{3}{2} ln x^{2} + 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x + 1))

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 2 x + 10 + \frac{2}{3} arctan (\frac{1}{3} (x + 1)) + C

\int x e n (x) d x

\int (x + \frac{1}{7 x}) d x

\frac{d}{d x} ((2 x^{6} + 5 x^{3})^{\frac{3}{5}})

n = 1 \sum \infty 1 + (\frac{1}{2})^{n}

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x

What is the integral of (3x+5)/(x^2+2x+10) ?
The integral of (3x+5)/(x^2+2x+10) is 3/2 ln|x^2+2x+10|+2/3 arctan(1/3 (x+1))+C