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

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

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

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

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

Solution steps

\int \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

Add a constant to the solution

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

\int ((x - 2) x + 4) d x

x \to 0 lim (x \cdot (e^{\frac{1}{x}} - 1))

y = arctan (2 t)

\frac{\partial}{\partial x} (2 x^{2} y^{4})

\int cos^{4} (2 x) d x

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