What is the integral of (ln(x)-1)/(ln^2(x)) ?

The integral of (ln(x)-1)/(ln^2(x)) is x/(ln(x))+C

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integral of (ln(x)-1)/(ln^2(x))

\int \frac{ln ( x ) - 1}{ln ^{2} ( x )} d x

\frac{x}{ln ( x )} + C

Solution steps

\int \frac{ln ( x ) - 1}{ln ^{2} ( x )} d x

Expand

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

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

\int \frac{1}{ln ( x )} d x = Ei (ln (x))

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

= Ei (ln (x)) - (- \frac{x}{ln ( x )} + Ei (ln (x)))

Simplify

= \frac{x}{ln ( x )}

Add a constant to the solution

= \frac{x}{ln ( x )} + C

What is the integral of (ln(x)-1)/(ln^2(x)) ?
The integral of (ln(x)-1)/(ln^2(x)) is x/(ln(x))+C