What is the integral of 2log_{10}(1/((2x+1))) ?

The integral of 2log_{10}(1/((2x+1))) is -2xlog_{10}(2x+1)+(2x+1-ln|2x+1|)/(ln(10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2log_{10}(1/((2x+1)))

integral

2 lo g_{10} (\frac{1}{( 2 x + 1 )})

- 2 x lo g_{10} (2 x + 1) + \frac{2 x + 1 - ln ∣ 2 x + 1 ∣}{ln ( 10 )} + C

Solution steps

\int 2 lo g_{10} (\frac{1}{2 x + 1}) d x

Take the constant out:

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

= 2 \cdot \int lo g_{10} (\frac{1}{2 x + 1}) d x

Apply Integration By Parts

: - x lo g_{10} (2 x + 1) - \int - \frac{2 x}{ln ( 10 ) ( 2 x + 1 )} d x

= 2 (- x lo g_{10} (2 x + 1) - \int - \frac{2 x}{ln ( 10 ) ( 2 x + 1 )} d x)

\int - \frac{2 x}{ln ( 10 ) ( 2 x + 1 )} d x = - \frac{2 x + 1 - ln ∣ 2 x + 1 ∣}{2 ln ( 10 )}

= 2 (- x lo g_{10} (2 x + 1) - (- \frac{2 x + 1 - ln ∣ 2 x + 1 ∣}{2 ln ( 10 )}))

Simplify

2 (- x lo g_{10} (2 x + 1) - (- \frac{2 x + 1 - ln ∣ 2 x + 1 ∣}{2 ln ( 10 )})) : - 2 x lo g_{10} (2 x + 1) + \frac{2 x + 1 - ln ∣ 2 x + 1 ∣}{ln ( 10 )}

= - 2 x lo g_{10} (2 x + 1) + \frac{2 x + 1 - ln ∣ 2 x + 1 ∣}{ln ( 10 )}

Add a constant to the solution

= - 2 x lo g_{10} (2 x + 1) + \frac{2 x + 1 - ln ∣ 2 x + 1 ∣}{ln ( 10 )} + C

e^{x} \cdot e^{2} x

\frac{1}{x w \cdot r t x}

2 t cosh (t^{2})

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

\frac{( 1 - x ^{8} )}{( 1 - x )}

What is the integral of 2log_{10}(1/((2x+1))) ?
The integral of 2log_{10}(1/((2x+1))) is -2xlog_{10}(2x+1)+(2x+1-ln|2x+1|)/(ln(10))+C