What is the integral of (5x^2)/((x-1)(x^2+9)) ?

The integral of (5x^2)/((x-1)(x^2+9)) is 5(1/10 ln|x-1|+1/20 (9ln|x^2+9|+6arctan(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 (5x^2)/((x-1)(x^2+9))

\int \frac{5 x ^{2}}{( x - 1 ) ( x ^{2} + 9 )} d x

5 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (9 ln x^{2} + 9 + 6 arctan (\frac{x}{3}))) + C

Solution steps

\int \frac{5 x ^{2}}{( x - 1 ) ( x ^{2} + 9 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{( x - 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 1 ) ( x ^{2} + 9 )} : \frac{1}{10 ( x - 1 )} + \frac{9 x + 9}{10 ( x ^{2} + 9 )}

= 5 \cdot \int \frac{1}{10 ( x - 1 )} + \frac{9 x + 9}{10 ( 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

= 5 (\int \frac{1}{10 ( x - 1 )} d x + \int \frac{9 x + 9}{10 ( x ^{2} + 9 )} d x)

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{9 x + 9}{10 ( x ^{2} + 9 )} d x = \frac{1}{20} (9 ln x^{2} + 9 + 6 arctan (\frac{x}{3}))

= 5 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (9 ln x^{2} + 9 + 6 arctan (\frac{x}{3})))

Add a constant to the solution

= 5 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{20} (9 ln x^{2} + 9 + 6 arctan (\frac{x}{3}))) + C

x \to 3 lim (x^{2} - 5 x)

\frac{d}{d s} (s^{2} y (s) - 1)

f (x) = x^{2} - 2 x + 1

\frac{d}{d x} (e^{- 0.4 x})

What is the integral of (5x^2)/((x-1)(x^2+9)) ?
The integral of (5x^2)/((x-1)(x^2+9)) is 5(1/10 ln|x-1|+1/20 (9ln|x^2+9|+6arctan(x/3)))+C