What is the integral of 1/(x^5+x^4) ?

The integral of 1/(x^5+x^4) is -ln|x|-1/x+1/(2x^2)-1/(3x^3)+ln|x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^5+x^4)

\int \frac{1}{x ^{5} + x ^{4}} d x

- ln ∣ x ∣ - \frac{1}{x} + \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} + ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{1}{x ^{5} + x ^{4}} d x

Factor

\frac{1}{x ^{5} + x ^{4}}

= \int \frac{1}{x ^{4} ( x + 1 )} d x

Take the partial fraction of

\frac{1}{x ^{4} ( x + 1 )} : - \frac{1}{x} + \frac{1}{x ^{2}} - \frac{1}{x ^{3}} + \frac{1}{x ^{4}} + \frac{1}{x + 1}

= \int - \frac{1}{x} + \frac{1}{x ^{2}} - \frac{1}{x ^{3}} + \frac{1}{x ^{4}} + \frac{1}{x + 1} 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}{x} d x + \int \frac{1}{x ^{2}} d x - \int \frac{1}{x ^{3}} d x + \int \frac{1}{x ^{4}} d x + \int \frac{1}{x + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

Simplify

= - ln ∣ x ∣ - \frac{1}{x} + \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} + ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ - \frac{1}{x} + \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} + ln ∣ x + 1 ∣ + C

\int \frac{( 8 \cdot x + 5 )}{( x ^{2} + 1 )} d x

\int csc^{4} (5 x) d x

\int 1 - x x^{2} d x

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

\int 7 cos (7 x) d x

What is the integral of 1/(x^5+x^4) ?
The integral of 1/(x^5+x^4) is -ln|x|-1/x+1/(2x^2)-1/(3x^3)+ln|x+1|+C