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

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

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

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

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

Solution steps

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

Simplify

= \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

\int_{- l n (4)}^{0} 23 e^{t} d t

\frac{\partial}{\partial z} (sin (x yz))

\frac{\partial}{\partial x} (e^{13 x y})

n = 1 \sum \infty \frac{cos ( n \cdot π )}{n ^{\frac{3}{8}}}

x \to 0 lim (\frac{sin ( π )}{x})

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