What is the integral of 1/(8x+1)-1/(e^x(8+e^{-x))^2} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(8x+1)-1/(e^x(8+e^{-x))^2}

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

\frac{1}{8} ln ∣ 8 x + 1 ∣ - \frac{1}{8 + e ^{- x}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{- 8 u + 1} + \frac{1}{e ^{- u} ( 8 + e ^{u} ) ^{2}} d u

= \int - \frac{1}{- 8 u + 1} + \frac{1}{e ^{- u} ( 8 + e ^{u} ) ^{2}} 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

= - \int \frac{1}{- 8 u + 1} d u + \int \frac{1}{e ^{- u} ( 8 + e ^{u} ) ^{2}} d u

\int \frac{1}{- 8 u + 1} d u = - \frac{1}{8} ln ∣ - 8 u + 1 ∣

\int \frac{1}{e ^{- u} ( 8 + e ^{u} ) ^{2}} d u = - \frac{1}{8 + e ^{u}}

= - (- \frac{1}{8} ln ∣ - 8 u + 1 ∣) - \frac{1}{8 + e ^{u}}

Substitute back

u = - x

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

Simplify

= \frac{1}{8} ln ∣ 8 x + 1 ∣ - \frac{1}{8 + e ^{- x}}

Add a constant to the solution

= \frac{1}{8} ln ∣ 8 x + 1 ∣ - \frac{1}{8 + e ^{- x}} + C

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

\int \frac{4 t ^{3} - t ^{2} + 36 t}{t ^{2} + 9} d t

\frac{\partial}{\partial x} (z - x y)

\frac{\partial}{\partial x} (sin (5 x + 6 y + z))

\int sin (5 x) cos (2 x) d x

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