What is the integral of (14x^2+8x-5)/(x^3+x^2) ?

The integral of (14x^2+8x-5)/(x^3+x^2) is 13ln|x|+5/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 (14x^2+8x-5)/(x^3+x^2)

\int \frac{14 x ^{2} + 8 x - 5}{x ^{3} + x ^{2}} d x

13 ln ∣ x ∣ + \frac{5}{x} + ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{14 x ^{2} + 8 x - 5}{x ^{3} + x ^{2}} d x

Take the partial fraction of

\frac{14 x ^{2} + 8 x - 5}{x ^{3} + x ^{2}} : \frac{13}{x} - \frac{5}{x ^{2}} + \frac{1}{x + 1}

= \int \frac{13}{x} - \frac{5}{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

= \int \frac{13}{x} d x - \int \frac{5}{x ^{2}} d x + \int \frac{1}{x + 1} d x

\int \frac{13}{x} d x = 13 ln ∣ x ∣

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

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

= 13 ln ∣ x ∣ - (- \frac{5}{x}) + ln ∣ x + 1 ∣

Simplify

= 13 ln ∣ x ∣ + \frac{5}{x} + ln ∣ x + 1 ∣

Add a constant to the solution

= 13 ln ∣ x ∣ + \frac{5}{x} + ln ∣ x + 1 ∣ + C

\frac{d}{d t} (\frac{e ^{- t} - e ^{- t} t}{e ^{t}})

n = 0 \sum \infty n (\frac{2}{3})^{n}

\frac{d}{d x} (cos (x) - x sin (x))

f (x) = x^{2} e^{- x^{2}}

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

What is the integral of (14x^2+8x-5)/(x^3+x^2) ?
The integral of (14x^2+8x-5)/(x^3+x^2) is 13ln|x|+5/x+ln|x+1|+C