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

The integral of (x+3)/(4x^4+4x^3+x^2) is -11ln|x|-3/x+11ln|2x+1|-5/(2x+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 11 ln ∣ x ∣ - \frac{3}{x} + 11 ln ∣ 2 x + 1 ∣ - \frac{5}{2 x + 1} + C

Solution steps

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

Take the partial fraction of

\frac{x + 3}{4 x ^{4} + 4 x ^{3} + x ^{2}} : - \frac{11}{x} + \frac{3}{x ^{2}} + \frac{22}{2 x + 1} + \frac{10}{( 2 x + 1 ) ^{2}}

= \int - \frac{11}{x} + \frac{3}{x ^{2}} + \frac{22}{2 x + 1} + \frac{10}{( 2 x + 1 ) ^{2}} 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{11}{x} d x + \int \frac{3}{x ^{2}} d x + \int \frac{22}{2 x + 1} d x + \int \frac{10}{( 2 x + 1 ) ^{2}} d x

\int \frac{11}{x} d x = 11 ln ∣ x ∣

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

\int \frac{22}{2 x + 1} d x = 11 ln ∣ 2 x + 1 ∣

\int \frac{10}{( 2 x + 1 ) ^{2}} d x = - \frac{5}{2 x + 1}

= - 11 ln ∣ x ∣ - \frac{3}{x} + 11 ln ∣ 2 x + 1 ∣ - \frac{5}{2 x + 1}

Add a constant to the solution

= - 11 ln ∣ x ∣ - \frac{3}{x} + 11 ln ∣ 2 x + 1 ∣ - \frac{5}{2 x + 1} + C

\int \frac{2 x + 16}{x ( x ^{2} - 25 )} d x

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

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

\int \frac{( y - 3 )}{- 24 + 14 y - y ^{2}} d y

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

What is the integral of (x+3)/(4x^4+4x^3+x^2) ?
The integral of (x+3)/(4x^4+4x^3+x^2) is -11ln|x|-3/x+11ln|2x+1|-5/(2x+1)+C