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

The integral of 1/((1+log_{10)(x))} is ln(10) 1/10 \Ei(ln(10^{1+log_{10}(x)}))+C

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integral of 1/((1+log_{10)(x))}

integral

\frac{1}{( 1 + lo g _{10} ( x ) )}

ln (10) \frac{1}{10} Ei (ln (1 0^{1 + l o g_{10} (x)})) + C

Solution steps

\int \frac{1}{1 + lo g _{10} ( x )} d x

Apply u-substitution

: \int \frac{ln ( 10 ) \cdot 1 0 ^{u - 1}}{u} d u

= \int \frac{ln ( 10 ) \cdot 1 0 ^{u - 1}}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= ln (10) \cdot \int \frac{1 0 ^{u - 1}}{u} d u

Simplify

1 0^{u - 1} : 1 0^{u} \cdot 1 0^{- 1}

= ln (10) \cdot \int \frac{1 0 ^{u} \cdot 1 0 ^{- 1}}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= ln (10) \cdot 1 0^{- 1} \cdot \int \frac{1 0 ^{u}}{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= ln (10) \frac{1}{10} \cdot \int \frac{1 0 ^{u}}{u} d u

Apply u-substitution

: \int \frac{1}{ln ( v )} d v

= ln (10) \frac{1}{10} \cdot \int \frac{1}{ln ( v )} d v

Apply u-substitution

: \int \frac{e ^{w}}{w} d w

= ln (10) \frac{1}{10} \cdot \int \frac{e ^{w}}{w} d w

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= ln (10) \frac{1}{10} Ei (w)

Substitute back

= ln (10) \frac{1}{10} Ei (ln (1 0^{1 + l o g_{10} (x)}))

Add a constant to the solution

= ln (10) \frac{1}{10} Ei (ln (1 0^{1 + l o g_{10} (x)})) + C

t^{- 1}

- e^{y} - 2

\frac{1}{( t ^{2} - 10 t + 1 )}

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

lo g_{10} (x^{2} + 4)

What is the integral of 1/((1+log_{10)(x))} ?
The integral of 1/((1+log_{10)(x))} is ln(10) 1/10 \Ei(ln(10^{1+log_{10}(x)}))+C