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

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

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

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

\frac{- ln ( 2 ) - 1}{2} + 1

Decimal Notation

0.15342 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Since the denominators are equal, combine the fractions:

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

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

Compute the boundaries:

\frac{- ln ( 2 ) - 1}{2} + 1

= \frac{- ln ( 2 ) - 1}{2} + 1

\int \frac{sin ( x )}{1 - sin ( x )} d x

\int 7 x^{6} - \frac{6}{x ^{6}} d x

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h \to 0 lim (\frac{( a + h ) - ( a )}{h})

n = 1 \sum \infty (5)^{- n}

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