What is the integral of 1/(sqrt(y^2+1)) ?

The integral of 1/(sqrt(y^2+1)) is ln|sqrt(y^2+1)+y|+C

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integral of 1/(sqrt(y^2+1))

\int \frac{1}{y ^{2} + 1} d y

ln y^{2} + 1 + y + C

Solution steps

\int \frac{1}{y ^{2} + 1} d y

Use the common integral:

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

= ln y^{2} + 1 + y

Add a constant to the solution

= ln y^{2} + 1 + y + C

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What is the integral of 1/(sqrt(y^2+1)) ?
The integral of 1/(sqrt(y^2+1)) is ln|sqrt(y^2+1)+y|+C