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

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

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

integral

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{1}{y ^{3}}

x^{2} - a^{2}

\frac{s}{( s ^{2} + 36 )}

\frac{- 2}{5 t + 4}

(x)^{5} sin (x)

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