What is the integral of (10)/((3x-3)(x^2+9)) ?

The integral of (10)/((3x-3)(x^2+9)) is 10(1/30 ln|x-1|+1/30 (-1/2 ln|x^2+9|-1/3 arctan(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 (10)/((3x-3)(x^2+9))

\int \frac{10}{( 3 x - 3 ) ( x ^{2} + 9 )} d x

10 (\frac{1}{30} ln ∣ x - 1 ∣ + \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) + C

Solution steps

\int \frac{10}{( 3 x - 3 ) ( x ^{2} + 9 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int \frac{1}{( 3 x - 3 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{1}{( 3 x - 3 ) ( x ^{2} + 9 )} : \frac{1}{30 ( x - 1 )} + \frac{- x - 1}{30 ( x ^{2} + 9 )}

= 10 \cdot \int \frac{1}{30 ( x - 1 )} + \frac{- x - 1}{30 ( x ^{2} + 9 )} 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

= 10 (\int \frac{1}{30 ( x - 1 )} d x + \int \frac{- x - 1}{30 ( x ^{2} + 9 )} d x)

\int \frac{1}{30 ( x - 1 )} d x = \frac{1}{30} ln ∣ x - 1 ∣

\int \frac{- x - 1}{30 ( x ^{2} + 9 )} d x = \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))

= 10 (\frac{1}{30} ln ∣ x - 1 ∣ + \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})))

Add a constant to the solution

= 10 (\frac{1}{30} ln ∣ x - 1 ∣ + \frac{1}{30} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) + C

\int \frac{1}{w ln ( w )} d w

\int cos^{2} (y) sin (y) d y

\int (arccos (x) - ln (x)) d x

\int (\frac{1}{2} x - 2)^{4} d x

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

What is the integral of (10)/((3x-3)(x^2+9)) ?
The integral of (10)/((3x-3)(x^2+9)) is 10(1/30 ln|x-1|+1/30 (-1/2 ln|x^2+9|-1/3 arctan(x/3)))+C