What is the integral from 1 to 2 of 1/(sqrt(x^2+1)) ?

The integral from 1 to 2 of 1/(sqrt(x^2+1)) is ln(sqrt(5)+2)-ln(sqrt(2)+1)

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integral from 1 to 2 of 1/(sqrt(x^2+1))

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

ln (5 + 2) - ln (2 + 1)

Decimal

0.56226 \dots

Solution steps

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

Use the common integral:

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

= [ln x^{2} + 1 + x]_{1}^{2}

Compute the boundaries:

ln (5 + 2) - ln (2 + 1)

= ln (5 + 2) - ln (2 + 1)

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What is the integral from 1 to 2 of 1/(sqrt(x^2+1)) ?
The integral from 1 to 2 of 1/(sqrt(x^2+1)) is ln(sqrt(5)+2)-ln(sqrt(2)+1)