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

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

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

integral

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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e^{6} x cos (e^{3} x)

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