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

The integral of (ln(y))/(log_{10)(y)} is 1/(log_{10)(e)}y+C

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integral of (ln(y))/(log_{10)(y)}

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

\frac{1}{lo g _{10} ( e )} y + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{lo g _{10} ( e )} d u

= \int \frac{e ^{u}}{lo g _{10} ( e )} d u

Take the constant out:

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

= \frac{1}{lo g _{10} ( e )} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{lo g _{10} ( e )} e^{u}

Substitute back

u = ln (y)

= \frac{1}{lo g _{10} ( e )} e^{l n (y)}

Apply log rule:

a^{l o g_{a} (b)} = b

= \frac{1}{lo g _{10} ( e )} y

Add a constant to the solution

= \frac{1}{lo g _{10} ( e )} y + C

f (x) = x^{4} - 17 x^{2} + 16

\int π (\frac{sin ( x )}{x})^{2} d x

x \to \frac{π}{2} lim (3 sin (x))

\int_{- 1}^{1} \frac{5 x}{( x + 4 ) ^{2}} d x

y = ∣ x - 6 ∣, y = \frac{x}{2}

What is the integral of (ln(y))/(log_{10)(y)} ?
The integral of (ln(y))/(log_{10)(y)} is 1/(log_{10)(e)}y+C