What is the integral of (9sqrt(x^2-1))/x ?

The integral of (9sqrt(x^2-1))/x is 9(-arctan(sqrt(x^2-1))+sqrt(x^2-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (9sqrt(x^2-1))/x

\int \frac{9 x ^{2} - 1}{x} d x

9 (- arctan (x^{2} - 1) + x^{2} - 1) + C

Solution steps

\int \frac{9 x ^{2} - 1}{x} d x

Take the constant out:

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

= 9 \cdot \int \frac{x ^{2} - 1}{x} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} + 1} d u

= 9 \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= 9 \cdot \int - \frac{1}{u ^{2} + 1} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

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

\int 1 d u = u

= 9 (- arctan (u) + u)

Substitute back

u = x^{2} - 1

= 9 (- arctan (x^{2} - 1) + x^{2} - 1)

Add a constant to the solution

= 9 (- arctan (x^{2} - 1) + x^{2} - 1) + C

\int \frac{6 t}{8 - 3 t ^{2}} d t

\int \frac{2}{x ^{2} ( x - 5 )} d x

\int \frac{( 5 x - 4 )}{x ^{3} + 4 x ^{2}} d x

\int x^{3} cos (1 + x^{2}) d x

\int sin (x) cos (x) e^{- s i n (x)} d x

What is the integral of (9sqrt(x^2-1))/x ?
The integral of (9sqrt(x^2-1))/x is 9(-arctan(sqrt(x^2-1))+sqrt(x^2-1))+C