What is the integral of (3x^3+12x+1)/((x^2+4)^2) ?

The integral of (3x^3+12x+1)/((x^2+4)^2) is 3/2 ln|x^2+4|+1/16 arctan(x/2)+1/32 sin(2arctan(x/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x^3+12x+1)/((x^2+4)^2)

\int \frac{3 x ^{3} + 12 x + 1}{( x ^{2} + 4 ) ^{2}} d x

\frac{3}{2} ln x^{2} + 4 + \frac{1}{16} arctan (\frac{x}{2}) + \frac{1}{32} sin (2 arctan (\frac{x}{2})) + C

Solution steps

\int \frac{3 x ^{3} + 12 x + 1}{( x ^{2} + 4 ) ^{2}} d x

Expand

\frac{3 x ^{3} + 12 x + 1}{( x ^{2} + 4 ) ^{2}} : \frac{3 x ^{3}}{( x ^{2} + 4 ) ^{2}} + \frac{12 x}{( x ^{2} + 4 ) ^{2}} + \frac{1}{( x ^{2} + 4 ) ^{2}}

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{3 x ^{3}}{( x ^{2} + 4 ) ^{2}} d x + \int \frac{12 x}{( x ^{2} + 4 ) ^{2}} d x + \int \frac{1}{( x ^{2} + 4 ) ^{2}} d x

\int \frac{3 x ^{3}}{( x ^{2} + 4 ) ^{2}} d x = \frac{3}{2} (ln x^{2} + 4 + \frac{4}{x ^{2} + 4})

\int \frac{12 x}{( x ^{2} + 4 ) ^{2}} d x = - \frac{6}{x ^{2} + 4}

\int \frac{1}{( x ^{2} + 4 ) ^{2}} d x = \frac{1}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

= \frac{3}{2} (ln x^{2} + 4 + \frac{4}{x ^{2} + 4}) - \frac{6}{x ^{2} + 4} + \frac{1}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

Simplify

\frac{3}{2} (ln x^{2} + 4 + \frac{4}{x ^{2} + 4}) - \frac{6}{x ^{2} + 4} + \frac{1}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) : \frac{3}{2} ln x^{2} + 4 + \frac{1}{16} arctan (\frac{x}{2}) + \frac{1}{32} sin (2 arctan (\frac{x}{2}))

= \frac{3}{2} ln x^{2} + 4 + \frac{1}{16} arctan (\frac{x}{2}) + \frac{1}{32} sin (2 arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 4 + \frac{1}{16} arctan (\frac{x}{2}) + \frac{1}{32} sin (2 arctan (\frac{x}{2})) + C

What is the integral of (3x^3+12x+1)/((x^2+4)^2) ?
The integral of (3x^3+12x+1)/((x^2+4)^2) is 3/2 ln|x^2+4|+1/16 arctan(x/2)+1/32 sin(2arctan(x/2))+C