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

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

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integral from 1 to x^2 of 1/t

\int_{1}^{x^{2}} \frac{1}{t} d t

2 ln (x)

Solution steps

\int_{1}^{x^{2}} \frac{1}{t} d t

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= [ln ∣ t ∣]_{1}^{x^{2}}

Compute the boundaries:

2 ln (x)

= 2 ln (x)

\int (\frac{x ^{4} - 3 x ^{2} + 5}{x ^{4}}) d x

x^{2} \frac{d y}{d x} - x = 1 + y + x y

f (x) = 12 x

\int e^{x^{2}} 2 x d x

\frac{1}{1 - x}, 9

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