What is the integral of (x-3)/(x^4-5x^2+4) ?

The integral of (x-3)/(x^4-5x^2+4) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-3)/(x^4-5x^2+4)

\int \frac{x - 3}{x ^{4} - 5 x ^{2} + 4} d x

- \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1) - 3 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{x - 3}{x ^{4} - 5 x ^{2} + 4} d x

Expand

\frac{x - 3}{x ^{4} - 5 x ^{2} + 4} : \frac{x}{x ^{4} - 5 x ^{2} + 4} - \frac{3}{x ^{4} - 5 x ^{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{x}{x ^{4} - 5 x ^{2} + 4} d x - \int \frac{3}{x ^{4} - 5 x ^{2} + 4} d x

\int \frac{x}{x ^{4} - 5 x ^{2} + 4} d x = - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1)

\int \frac{3}{x ^{4} - 5 x ^{2} + 4} d x = 3 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣)

= - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1) - 3 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣)

Add a constant to the solution

= - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1) - 3 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣) + C

\int \frac{x - 56}{x ^{2} + x - 56} d x

\int \frac{sin ( 2 x )}{9 + cos ^{2} ( 2 x )} d x

\int e^{2 x} y d x

\int \frac{x ^{2}}{2 x - x} d x

\int \frac{e ^{x}}{e ^{2 x} + 9 e ^{x} + 20} d x

What is the integral of (x-3)/(x^4-5x^2+4) ?
The integral of (x-3)/(x^4-5x^2+4) is