What is the integral of ((e^xsqrt(e^x-1)))/(e^x+8) ?

The integral of ((e^xsqrt(e^x-1)))/(e^x+8) is 6(-arctan(1/3 sqrt(e^x-1))+1/3 sqrt(e^x-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((e^xsqrt(e^x-1)))/(e^x+8)

\int \frac{( e ^{x} e ^{x} - 1 )}{e ^{x} + 8} d x

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

Solution steps

\int \frac{e ^{x} e ^{x} - 1}{e ^{x} + 8} d x

Apply u-substitution

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

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

Take the constant out:

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

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

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

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

Apply Integral Substitution

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

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}{v ^{2} + 1} + 1 d v

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}{v ^{2} + 1} d v + \int 1 d v)

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

\int 1 d v = v

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( y )} d y

\int 0.09 e^{0.04 x} d x

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

\int \frac{21 x ^{2}}{sec ( x ^{3} - 3 )} d x

\int x (e^{x^{2}} + 12) d x

What is the integral of ((e^xsqrt(e^x-1)))/(e^x+8) ?
The integral of ((e^xsqrt(e^x-1)))/(e^x+8) is 6(-arctan(1/3 sqrt(e^x-1))+1/3 sqrt(e^x-1))+C