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

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

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integral of (log_{10}(log_{10}(x)))/x

\int \frac{lo g _{10} ( lo g _{10} ( x ) )}{x} d x

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

Solution steps

\int \frac{lo g _{10} ( lo g _{10} ( x ) )}{x} d x

Apply u-substitution

: \int ln (u) d u

= \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= u ln (u) - \int 1 d u

\int 1 d u = u

= u ln (u) - u

Substitute back

u = lo g_{10} (x)

= lo g_{10} (x) ln (lo g_{10} (x)) - lo g_{10} (x)

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{2}}{x ^{3} + 1})

\frac{\partial}{\partial x} (e^{- y} + e^{- x})

f (x) = 4 sin (4 x)

\frac{d}{d x} (ln (3 + x^{2}))

x^{^{'}} + 6 x = 0

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