What is the integral of ((x+7))/((x^2-6x+34)) ?

The integral of ((x+7))/((x^2-6x+34)) is 1/2 ln|x^2-6x+34|+2arctan(1/5 (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 ((x+7))/((x^2-6x+34))

integral

\frac{( x + 7 )}{( x ^{2} - 6 x + 34 )}

\frac{1}{2} ln x^{2} - 6 x + 34 + 2 arctan (\frac{1}{5} (x - 3)) + C

Solution steps

\int \frac{x + 7}{x ^{2} - 6 x + 34} d x

Complete the square

x^{2} - 6 x + 34 : (x - 3)^{2} + 25

= \int \frac{x + 7}{( x - 3 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{u + 10}{u ^{2} + 25} d u

= \int \frac{u + 10}{u ^{2} + 25} d u

Expand

\frac{u + 10}{u ^{2} + 25} : \frac{u}{u ^{2} + 25} + \frac{10}{u ^{2} + 25}

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{u}{u ^{2} + 25} d u + \int \frac{10}{u ^{2} + 25} d u

\int \frac{u}{u ^{2} + 25} d u = \frac{1}{2} ln u^{2} + 25

\int \frac{10}{u ^{2} + 25} d u = 2 arctan (\frac{u}{5})

= \frac{1}{2} ln u^{2} + 25 + 2 arctan (\frac{u}{5})

Substitute back

u = x - 3

= \frac{1}{2} ln (x - 3)^{2} + 25 + 2 arctan (\frac{x - 3}{5})

Simplify

\frac{1}{2} ln (x - 3)^{2} + 25 + 2 arctan (\frac{x - 3}{5}) : \frac{1}{2} ln x^{2} - 6 x + 34 + 2 arctan (\frac{1}{5} (x - 3))

= \frac{1}{2} ln x^{2} - 6 x + 34 + 2 arctan (\frac{1}{5} (x - 3))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 6 x + 34 + 2 arctan (\frac{1}{5} (x - 3)) + C

(1 - x)^{n}

\frac{1}{2 sin ( 2 x )}

(a^{2} + x^{2})^{\frac{1}{2}}

x^{4} (1 - x)^{5}

\frac{x ^{3}}{( x ^{2} + x + 2 )}

What is the integral of ((x+7))/((x^2-6x+34)) ?
The integral of ((x+7))/((x^2-6x+34)) is 1/2 ln|x^2-6x+34|+2arctan(1/5 (x-3))+C