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

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

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integral of log_{10}(x^2)

integral

lo g_{10} (x^{2})

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

Solution steps

\int lo g_{10} (x^{2}) d x

Apply Integration By Parts

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

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

\int \frac{2}{ln ( 10 )} d x = \frac{2}{ln ( 10 )} x

= x lo g_{10} (x^{2}) - \frac{2}{ln ( 10 )} x

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

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

Add a constant to the solution

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

\frac{( x ^{5} + 1 )}{( x - 1 )}

\frac{e ^{x^{2}}}{x}

\frac{x ^{2} - 12 x + 13}{x}

\frac{( x + 3 )}{( x - 1 )}

\frac{1}{( e ^{x^{2}} )}

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