What is the integral of (x+3)/(x^2+4x+13) ?

The integral of (x+3)/(x^2+4x+13) is 1/2 ln|x^2+4x+13|+1/3 arctan((x+2)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+3)/(x^2+4x+13)

\int \frac{x + 3}{x ^{2} + 4 x + 13} d x

\frac{1}{2} ln x^{2} + 4 x + 13 + \frac{1}{3} arctan (\frac{x + 2}{3}) + C

Solution steps

\int \frac{x + 3}{x ^{2} + 4 x + 13} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 2 u + 10} d u

= \int \frac{u}{u ^{2} - 2 u + 10} d u

Complete the square

u^{2} - 2 u + 10 : (u - 1)^{2} + 9

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

Apply u-substitution

: \int \frac{v + 1}{v ^{2} + 9} d v

= \int \frac{v + 1}{v ^{2} + 9} d v

Expand

\frac{v + 1}{v ^{2} + 9} : \frac{v}{v ^{2} + 9} + \frac{1}{v ^{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{v}{v ^{2} + 9} d v + \int \frac{1}{v ^{2} + 9} d v

\int \frac{v}{v ^{2} + 9} d v = \frac{1}{2} ln v^{2} + 9

\int \frac{1}{v ^{2} + 9} d v = \frac{1}{3} arctan (\frac{v}{3})

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

Substitute back

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

Simplify

\frac{1}{2} ln (x + 3 - 1)^{2} + 9 + \frac{1}{3} arctan (\frac{x + 3 - 1}{3}) : \frac{1}{2} ln x^{2} + 4 x + 13 + \frac{1}{3} arctan (\frac{x + 2}{3})

= \frac{1}{2} ln x^{2} + 4 x + 13 + \frac{1}{3} arctan (\frac{x + 2}{3})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 13 + \frac{1}{3} arctan (\frac{x + 2}{3}) + C

\int \frac{1}{4 x ^{2} + 12 x + 25} d x

\int y^{5} y^{3} - 2 d y

\int sin (\frac{v}{3}) cos (\frac{v}{3}) d v

\int (\frac{2 x ^{4} - x}{x ^{3}}) d x

\int \frac{sin ( x )}{1 + cos ( x )} d x

What is the integral of (x+3)/(x^2+4x+13) ?
The integral of (x+3)/(x^2+4x+13) is 1/2 ln|x^2+4x+13|+1/3 arctan((x+2)/3)+C