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

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

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

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

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

Solution steps

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

Use the common integral:

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

= ln t^{2} + 1 + t

Add a constant to the solution

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

- sec (x) + 2 sec^{3} (x)

\frac{x}{x ^{2} + 25}

\frac{d}{d x} (arctan (a x))

\int (2 x^{- 3}) d x

\frac{d}{d x} (\frac{( x ^{2} - 1 ) ( x + 1 )}{x ^{2} - 2 x + 1})

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