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

The integral of (ln(x)+1)/(x^2) is -(ln(x))/x-2/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 )}{x} - \frac{2}{x} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

= - \frac{ln ( x )}{x} - \frac{2}{x}

Add a constant to the solution

= - \frac{ln ( x )}{x} - \frac{2}{x} + C

f (x) = 5 e^{x} cos (x)

\int_{- 6}^{x} 36 - t^{2} d t

(1 + x)^{\frac{1}{2}}, 0

y^{^{'}} = (y + 5) (x + 2), y (0) = - 1

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