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

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

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

integral

lo g_{10} (1 + x^{2})

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

x^{2} - x + 1

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

\frac{( 3 x ^{2} - 1 )}{( x ^{3} - x )}

\frac{x + 1}{x ^{2} + x + 6}

3 e^{3} x

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