What is the integral of 1/((x+8)(x^2+16)) ?

The integral of 1/((x+8)(x^2+16)) is 1/80 ln|x+8|+1/80 (-1/2 ln|(x^2)/(16)+1|+2arctan(x/4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((x+8)(x^2+16))

\int \frac{1}{( x + 8 ) ( x ^{2} + 16 )} d x

\frac{1}{80} ln ∣ x + 8 ∣ + \frac{1}{80} (- \frac{1}{2} ln \frac{x ^{2}}{16} + 1 + 2 arctan (\frac{x}{4})) + C

Solution steps

\int \frac{1}{( x + 8 ) ( x ^{2} + 16 )} d x

Take the partial fraction of

\frac{1}{( x + 8 ) ( x ^{2} + 16 )} : \frac{1}{80 ( x + 8 )} + \frac{- x + 8}{80 ( x ^{2} + 16 )}

= \int \frac{1}{80 ( x + 8 )} + \frac{- x + 8}{80 ( x ^{2} + 16 )} d x

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{1}{80 ( x + 8 )} d x + \int \frac{- x + 8}{80 ( x ^{2} + 16 )} d x

\int \frac{1}{80 ( x + 8 )} d x = \frac{1}{80} ln ∣ x + 8 ∣

\int \frac{- x + 8}{80 ( x ^{2} + 16 )} d x = \frac{1}{80} (- \frac{1}{2} ln \frac{x ^{2}}{16} + 1 + 2 arctan (\frac{x}{4}))

= \frac{1}{80} ln ∣ x + 8 ∣ + \frac{1}{80} (- \frac{1}{2} ln \frac{x ^{2}}{16} + 1 + 2 arctan (\frac{x}{4}))

Add a constant to the solution

= \frac{1}{80} ln ∣ x + 8 ∣ + \frac{1}{80} (- \frac{1}{2} ln \frac{x ^{2}}{16} + 1 + 2 arctan (\frac{x}{4})) + C

\int sin (4 x + 5) d x

\int sin^{2} (t) cos^{4} (t) d t

\frac{d}{d x} (\frac{2 x}{( x + 1 )})

\int \frac{2 x + 1}{x ^{2} + 2 x + 5} d x

\int \frac{1}{2} e^{2 x} d x

What is the integral of 1/((x+8)(x^2+16)) ?
The integral of 1/((x+8)(x^2+16)) is 1/80 ln|x+8|+1/80 (-1/2 ln|(x^2)/(16)+1|+2arctan(x/4))+C