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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Expand

\frac{2 x - 1}{x ^{2} + 4} : \frac{2 x}{x ^{2} + 4} - \frac{1}{x ^{2} + 4}

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

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

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

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

Add a constant to the solution

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

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