What is the integral from 0 to 1 of 1/(sqrt(x^2+1)) ?

The integral from 0 to 1 of 1/(sqrt(x^2+1)) is ln(1+sqrt(2))

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integral from 0 to 1 of 1/(sqrt(x^2+1))

\int_{0}^{1} \frac{1}{x ^{2} + 1} d x

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

\int_{0}^{1} \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]_{0}^{1}

Compute the boundaries:

ln (1 + 2)

= ln (1 + 2)

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

\int_{0}^{1} 2 cos (\frac{π x}{2}) d x

\int_{0}^{2 π} 0 d x

\int_{0}^{1} (1 + x^{2})^{- \frac{3}{2}} d x

\int_{a}^{x} f (t) d t

What is the integral from 0 to 1 of 1/(sqrt(x^2+1)) ?
The integral from 0 to 1 of 1/(sqrt(x^2+1)) is ln(1+sqrt(2))