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

The integral of (x^3+4x+3)/(x^2+2) is 1/2 x^2+1+ln|x^2+2|+3/(sqrt(2))arctan(x/(sqrt(2)))+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+3)/(x^2+2)

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

\frac{1}{2} x^{2} + 1 + ln x^{2} + 2 + \frac{3}{2} arctan (\frac{x}{2}) + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

= \frac{1}{2} x^{2} + 1 + ln x^{2} + 2 + \frac{3}{2} arctan (\frac{x}{2})

Add a constant to the solution

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

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

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

\int \frac{e ^{5 x}}{3 x} d x

\int \frac{2}{y ^{4}} d y

\int \frac{1}{4 - x ^{2} ^{3}} d x

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