What is the integral of (8sqrt(x^2-1))/(x^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8sqrt(x^2-1))/(x^2)

\int \frac{8 x ^{2} - 1}{x ^{2}} d x

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

Solution steps

\int \frac{8 x ^{2} - 1}{x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{x ^{2} - 1}{x ^{2}} d x

Apply Integration By Parts

: - \frac{x ^{2} - 1}{x} - \int - \frac{1}{x ^{2} - 1} d x

= 8 (- \frac{x ^{2} - 1}{x} - \int - \frac{1}{x ^{2} - 1} d x)

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

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

Simplify

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

Add a constant to the solution

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

f (x) \frac{7}{x}, at a = 3

\int \frac{e ^{9 x}}{e ^{9 x} + 8} d x

y = e^{x}, y = x e^{x}, x = 0

(\frac{- ln ( 1 + 6 x )}{x})^{^{'}}

\int (2 x - 1) e^{x} d x

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