What is the integral of (x+17)/(x^2+28x+200) ?

The integral of (x+17)/(x^2+28x+200) is 1/2 ln|x^2+28x+200|+3/2 arctan((x+14)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+17)/(x^2+28x+200)

\int \frac{x + 17}{x ^{2} + 28 x + 200} d x

\frac{1}{2} ln x^{2} + 28 x + 200 + \frac{3}{2} arctan (\frac{x + 14}{2}) + C

Solution steps

\int \frac{x + 17}{x ^{2} + 28 x + 200} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 6 u + 13} d u

= \int \frac{u}{u ^{2} - 6 u + 13} d u

Complete the square

u^{2} - 6 u + 13 : (u - 3)^{2} + 4

= \int \frac{u}{( u - 3 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{v + 3}{v ^{2} + 4} d v

= \int \frac{v + 3}{v ^{2} + 4} d v

Expand

\frac{v + 3}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{3}{v ^{2} + 4}

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} + 4} d v + \int \frac{3}{v ^{2} + 4} d v

\int \frac{v}{v ^{2} + 4} d v = \frac{1}{2} ln v^{2} + 4

\int \frac{3}{v ^{2} + 4} d v = \frac{3}{2} arctan (\frac{v}{2})

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

Substitute back

= \frac{1}{2} ln (x + 17 - 3)^{2} + 4 + \frac{3}{2} arctan (\frac{x + 17 - 3}{2})

Simplify

\frac{1}{2} ln (x + 17 - 3)^{2} + 4 + \frac{3}{2} arctan (\frac{x + 17 - 3}{2}) : \frac{1}{2} ln x^{2} + 28 x + 200 + \frac{3}{2} arctan (\frac{x + 14}{2})

= \frac{1}{2} ln x^{2} + 28 x + 200 + \frac{3}{2} arctan (\frac{x + 14}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 28 x + 200 + \frac{3}{2} arctan (\frac{x + 14}{2}) + C

\int_{0}^{π} cos^{4} (x) d x

\frac{d}{d x} (1 - sin (x) cos (x))

x \to \infty lim (x^{0})

\int t csc^{2} (\frac{3 t ^{2}}{2}) d t

n = 4 \sum \infty \frac{1}{n ^{2}}

What is the integral of (x+17)/(x^2+28x+200) ?
The integral of (x+17)/(x^2+28x+200) is 1/2 ln|x^2+28x+200|+3/2 arctan((x+14)/2)+C