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

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

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integral of log_{10}(x)+1

integral

lo g_{10} (x) + 1

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d x = x

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

Add similar elements:

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

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

Add a constant to the solution

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

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

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

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

\frac{2}{t}

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

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