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

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

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

integral

\frac{lo g _{10} ( ( 1 + x ) ^{2} )}{x ^{2}}

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

Solution steps

\int \frac{lo g _{10} ( ( 1 + x ) ^{2} )}{x ^{2}} d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

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