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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\frac{1}{y ^{4}}

\int 40

\frac{d}{d x} (tan (x + \frac{π}{4}))

\frac{s + 1}{s ^{2} + s + 3}

f (x) = e^{- 7 x}

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