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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

= \int \frac{u ^{2} + 1}{2 u ^{3}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u ^{2} + 1}{u ^{3}} d u

Expand

\frac{u ^{2} + 1}{u ^{3}} : \frac{1}{u} + \frac{1}{u ^{3}}

= \frac{1}{2} \cdot \int \frac{1}{u} + \frac{1}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= \frac{1}{2} (ln ∣ u ∣ - \frac{1}{2 u ^{2}})

Substitute back

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

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

Add a constant to the solution

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

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