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

The integral of ((x+1))/(2+4x^2) is 1/8 ln|2+4x^2|+1/(2sqrt(2))arctan(sqrt(2)x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{8} ln 2 + 4 x^{2} + \frac{1}{2 2} arctan (2 x) + C

Solution steps

\int \frac{x + 1}{2 + 4 x ^{2}} d x

Expand

\frac{x + 1}{2 + 4 x ^{2}} : \frac{x}{2 + 4 x ^{2}} + \frac{1}{2 + 4 x ^{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}{2 + 4 x ^{2}} d x + \int \frac{1}{2 + 4 x ^{2}} d x

\int \frac{x}{2 + 4 x ^{2}} d x = \frac{1}{8} ln 2 + 4 x^{2}

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

= \frac{1}{8} ln 2 + 4 x^{2} + \frac{1}{2 2} arctan (2 x)

Add a constant to the solution

= \frac{1}{8} ln 2 + 4 x^{2} + \frac{1}{2 2} arctan (2 x) + C

\int (3 - 2 x)^{8} d x

\int x^{5} ln (2 x) d x

\int \frac{2 x ^{4} + 6 x ^{2} - 108}{x ^{2} + 3 x} d x

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

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

What is the integral of ((x+1))/(2+4x^2) ?
The integral of ((x+1))/(2+4x^2) is 1/8 ln|2+4x^2|+1/(2sqrt(2))arctan(sqrt(2)x)+C