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

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

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

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

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

Solution steps

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

Use the common integral:

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

= ln u^{2} + 1 + u

Add a constant to the solution

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

y = x^{2}, y = x + 6

\frac{d}{d x} (- ln (- x))

f (x) = \frac{e ^{x}}{2 x + 1}

y^{^{''}} + 11 y^{^{'}} + 30 y = 0

f (x, y) = x^{2} + y^{2}

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