What is the integral of (1+x^2)^{-1/2} ?

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

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integral of (1+x^2)^{-1/2}

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

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

Solution steps

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply Integration By Parts

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

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

\int - \frac{x ^{2}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x = - ln x + 1 + x^{2} + \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

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

Simplify

= ln x + x^{2} + 1

Add a constant to the solution

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

(- \frac{2}{7}, \frac{2}{3}), (\frac{5}{7}, - \frac{7}{3})

x \to 1 lim (\frac{x - 1}{x ^{n} - 1})

(x + y)^{2} d x + (2 x y + x^{2} - 4) d y = 0

\frac{d}{d x} (5 e^{x} + 3)

\frac{d}{d x} ((x^{2} + 4) e^{- 2 x})

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