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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply Integration By Parts

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

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

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

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

(x + 6)^{2}

\frac{d}{d x} (\frac{5}{x ^{5}} + \frac{3}{x ^{2}})

n = 1 \sum \infty 9 e^{- n}

\int_{- 1}^{3} x e^{x y} d y

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