What is the integral from 1 to t of (ln(x))/(x^2) ?

The integral from 1 to t of (ln(x))/(x^2) is (-ln(t)-1)/t+1

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integral from 1 to t of (ln(x))/(x^2)

\int_{1}^{t} \frac{ln ( x )}{x ^{2}} d x

\frac{- ln ( t ) - 1}{t} + 1

Solution steps

\int_{1}^{t} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{t}

= [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{t}

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

= [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{t}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= [\frac{- ln ( x ) - 1}{x}]_{1}^{t}

Compute the boundaries:

\frac{- ln ( t ) - 1}{t} + 1

= \frac{- ln ( t ) - 1}{t} + 1

\int_{0}^{e - 1} \frac{1}{x + 1} d x

\int_{0}^{0.1} \frac{1}{1 + x ^{4}} d x

\int_{0}^{1} (2 + e^{x})^{2} d x

\int_{- 3}^{1} (2 x + 1) (3 + x) (1 - x) d x

\int_{- 1}^{2} - x^{2} + 4 d x

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