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

The integral of x/(2x^2+3x+2) is -3/(2sqrt(7))arctan(1/(sqrt(7))(4x+3))+1/4 ln|16x^2+24x+16|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{3}{2 7} arctan (\frac{1}{7} (4 x + 3)) + \frac{1}{4} ln 16 x^{2} + 24 x + 16 + C

Solution steps

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

Apply Integration By Parts

: \frac{2}{7} x arctan (\frac{1}{7} (4 x + 3)) - \int \frac{2}{7} arctan (\frac{1}{7} (4 x + 3)) d x

= \frac{2}{7} x arctan (\frac{1}{7} (4 x + 3)) - \int \frac{2}{7} arctan (\frac{1}{7} (4 x + 3)) d x

\int \frac{2}{7} arctan (\frac{1}{7} (4 x + 3)) d x = \frac{2}{7} x arctan (\frac{1}{7} (4 x + 3)) + \frac{3}{2 7} arctan (\frac{1}{7} (4 x + 3)) - \frac{1}{4} ln 16 x^{2} + 24 x + 16

= \frac{2}{7} x arctan (\frac{1}{7} (4 x + 3)) - (\frac{2}{7} x arctan (\frac{1}{7} (4 x + 3)) + \frac{3}{2 7} arctan (\frac{1}{7} (4 x + 3)) - \frac{1}{4} ln 16 x^{2} + 24 x + 16)

Simplify

\frac{2}{7} x arctan (\frac{1}{7} (4 x + 3)) - (\frac{2}{7} x arctan (\frac{1}{7} (4 x + 3)) + \frac{3}{2 7} arctan (\frac{1}{7} (4 x + 3)) - \frac{1}{4} ln 16 x^{2} + 24 x + 16) : - \frac{3}{2 7} arctan (\frac{1}{7} (4 x + 3)) + \frac{1}{4} ln 16 x^{2} + 24 x + 16

= - \frac{3}{2 7} arctan (\frac{1}{7} (4 x + 3)) + \frac{1}{4} ln 16 x^{2} + 24 x + 16

Add a constant to the solution

= - \frac{3}{2 7} arctan (\frac{1}{7} (4 x + 3)) + \frac{1}{4} ln 16 x^{2} + 24 x + 16 + C

\int \frac{x ^{4}}{x ^{5} - 5} d x

\int \frac{6}{x ln ( 8 x )} d x

\int ln (x) x^{5} d x

\int (\frac{1}{x} - \frac{3}{16 x ^{2} + 1}) d x

\int (\frac{x}{x - 5}) d x

What is the integral of x/(2x^2+3x+2) ?
The integral of x/(2x^2+3x+2) is -3/(2sqrt(7))arctan(1/(sqrt(7))(4x+3))+1/4 ln|16x^2+24x+16|+C