What is the integral of (2x^3+6x+3)/((x+2)^2(x^2+1)) ?

The integral of (2x^3+6x+3)/((x+2)^2(x^2+1)) is 2ln|x+2|+5/(x+2)+arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^3+6x+3)/((x+2)^2(x^2+1))

\int \frac{2 x ^{3} + 6 x + 3}{( x + 2 ) ^{2} ( x ^{2} + 1 )} d x

2 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + arctan (x) + C

Solution steps

\int \frac{2 x ^{3} + 6 x + 3}{( x + 2 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{2 x ^{3} + 6 x + 3}{( x + 2 ) ^{2} ( x ^{2} + 1 )} : \frac{2}{x + 2} - \frac{5}{( x + 2 ) ^{2}} + \frac{1}{x ^{2} + 1}

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

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

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

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

= 2 ln ∣ x + 2 ∣ - (- \frac{5}{x + 2}) + arctan (x)

Simplify

= 2 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + arctan (x)

Add a constant to the solution

= 2 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + arctan (x) + C

\int (t^{2} + t) e^{t^{3}} + 3^{t} d t

x \to 0 lim (ln (x^{2}) (1 e^{x} + 1))

\frac{d}{d t} (3 cos^{3} (t))

3^{(- \frac{1}{10} t)}

\frac{- 5 e ^{(- 3 s)}}{( s ^{2} - 5 s + 6 )}

What is the integral of (2x^3+6x+3)/((x+2)^2(x^2+1)) ?
The integral of (2x^3+6x+3)/((x+2)^2(x^2+1)) is 2ln|x+2|+5/(x+2)+arctan(x)+C