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

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

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

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

Solution steps

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

Expand

\frac{x + 3}{x ^{2} - 4} : \frac{x}{x ^{2} - 4} + \frac{3}{x ^{2} - 4}

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

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

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

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

Add a constant to the solution

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

n = 1 \sum \infty 1 4^{n} x^{n} n!

\int_{0}^{t} \frac{1}{x} d x

\int cos (x) sec^{7} (x) d x

\int \frac{x ^{3} - 6 x + 5}{x} d x

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