What is the integral from 1 to 9 of 1/t ?

The integral from 1 to 9 of 1/t is 2ln(3)

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integral from 1 to 9 of 1/t

\int_{1}^{9} \frac{1}{t} d t

2 ln (3)

Decimal

2.19722 \dots

Solution steps

\int_{1}^{9} \frac{1}{t} d t

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= [ln ∣ t ∣]_{1}^{9}

Compute the boundaries:

2 ln (3)

= 2 ln (3)

\int_{0}^{π} sin (2) (x) d x

\int_{0}^{2} (2 x - x^{2})^{3} d x

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

\int_{4}^{9} \frac{1}{x - x} d x

\int_{3}^{5} \frac{1}{( x - 3 ) ^{\frac{3}{2}}} d x

What is the integral from 1 to 9 of 1/t ?
The integral from 1 to 9 of 1/t is 2ln(3)