What is the integral of sqrt(e^x-9) ?

The integral of sqrt(e^x-9) is 6(-arctan(1/3 sqrt(e^x-9))+1/3 sqrt(e^x-9))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(e^x-9)

\int e^{x} - 9 d x

6 (- arctan (\frac{1}{3} e^{x} - 9) + \frac{1}{3} e^{x} - 9) + C

Solution steps

\int e^{x} - 9 d x

Apply u-substitution

: \int \frac{u - 9}{u} d u

= \int \frac{u - 9}{u} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 9} d v

= \int \frac{2 v ^{2}}{v ^{2} + 9} d v

Take the constant out:

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

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

\frac{v ^{2}}{v ^{2} + 9} = - \frac{9}{v ^{2} + 9} + 1

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

Apply Integral Substitution

= 2 \cdot \int 3 (- \frac{1}{w ^{2} + 1} + 1) d w

Take the constant out:

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

= 2 \cdot 3 \cdot \int - \frac{1}{w ^{2} + 1} + 1 d w

Apply the Sum Rule:

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

= 2 \cdot 3 (- \int \frac{1}{w ^{2} + 1} d w + \int 1 d w)

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

\int 1 d w = w

= 2 \cdot 3 (- arctan (w) + w)

Substitute back

= 2 \cdot 3 (- arctan (\frac{e ^{x} - 9}{3}) + \frac{e ^{x} - 9}{3})

Simplify

2 \cdot 3 (- arctan (\frac{e ^{x} - 9}{3}) + \frac{e ^{x} - 9}{3}) : 6 (- arctan (\frac{1}{3} e^{x} - 9) + \frac{1}{3} e^{x} - 9)

= 6 (- arctan (\frac{1}{3} e^{x} - 9) + \frac{1}{3} e^{x} - 9)

Add a constant to the solution

= 6 (- arctan (\frac{1}{3} e^{x} - 9) + \frac{1}{3} e^{x} - 9) + C

\int \frac{cos ( 4 x )}{sin ( 2 x )} d x

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

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

\int sin (0.01 t) + t^{3} d t

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

What is the integral of sqrt(e^x-9) ?
The integral of sqrt(e^x-9) is 6(-arctan(1/3 sqrt(e^x-9))+1/3 sqrt(e^x-9))+C