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

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

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

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

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

Solution steps

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

Use the common integral:

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

= ln u^{2} + 1 + u

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{- 2 x}{( x + y ) ^{2}})

\int \frac{1}{( x ^{2} + y ^{2} ) ^{3}} d y

\int cos (t^{3}) d t

\frac{d}{d x} (ln (9) x^{6})

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