What is the integral of (13)/(x^3+x^2) ?

The integral of (13)/(x^3+x^2) is 13(-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 (13)/(x^3+x^2)

\int \frac{13}{x ^{3} + x ^{2}} d x

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

Solution steps

\int \frac{13}{x ^{3} + x ^{2}} d x

Take the constant out:

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

= 13 \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}

= 13 \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

= 13 (- \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 ∣

= 13 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣)

Add a constant to the solution

= 13 (- ln ∣ x ∣ - \frac{1}{x} + ln ∣ x + 1 ∣) + C

\int e^{3 - 6 x} d x

\int y^{2} csc^{2} (y^{3}) cot^{2} (y^{3}) d y

\int \frac{6 x}{( x ^{2} + 4 ) ^{2}} d x

\int e^{x} csc^{2} (e^{x}) d x

\int \frac{x - 3}{x ^{2} + 7 x + 10} d x

What is the integral of (13)/(x^3+x^2) ?
The integral of (13)/(x^3+x^2) is 13(-ln|x|-1/x+ln|x+1|)+C