What is the integral of (20)/(x^3+x^2) ?

The integral of (20)/(x^3+x^2) is 20(-ln|x|-1/x+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 (20)/(x^3+x^2)

\int \frac{20}{x ^{3} + x ^{2}} d x

20 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣) + C

Solution steps

\int \frac{20}{x ^{3} + x ^{2}} d x

Take the constant out:

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

= 20 \cdot \int \frac{1}{x ^{3} + x ^{2}} d x

Take the partial fraction of

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

= 20 \cdot \int - \frac{1}{x} + \frac{1}{x ^{2}} + \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

= 20 (- \int \frac{1}{x} d x + \int \frac{1}{x ^{2}} 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 + 1} d x = ln ∣ x + 1 ∣

= 20 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣)

Add a constant to the solution

= 20 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣) + C

\int x (x^{2} - 3)^{5} d x

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

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

\int (2 x + 5) x + 3 d x

\int sec^{3} (8 x) tan (8 x) d x

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