What is the integral of sqrt(x+\sqrt{x^2+2)} ?

The integral of sqrt(x+\sqrt{x^2+2)} is 1/3 (x+sqrt(x^2+2))^{3/2}-2/(sqrt(x+\sqrt{x^2+2))}+C

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integral of sqrt(x+\sqrt{x^2+2)}

\int x + x^{2} + 2 d x

\frac{1}{3} (x + x^{2} + 2)^{\frac{3}{2}} - \frac{2}{x + x ^{2} + 2} + C

Solution steps

\int x + x^{2} + 2 d x

Apply u-substitution

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

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

Expand

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

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

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

\int \frac{2}{u u} d u = - \frac{4}{u}

= \frac{1}{2} (\frac{2}{3} u^{\frac{3}{2}} - \frac{4}{u})

Substitute back

u = x + x^{2} + 2

= \frac{1}{2} (\frac{2}{3} (x + x^{2} + 2)^{\frac{3}{2}} - \frac{4}{x + x ^{2} + 2})

Simplify

\frac{1}{2} (\frac{2}{3} (x + x^{2} + 2)^{\frac{3}{2}} - \frac{4}{x + x ^{2} + 2}) : \frac{1}{3} (x + x^{2} + 2)^{\frac{3}{2}} - \frac{2}{x + x ^{2} + 2}

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

Add a constant to the solution

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

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

\frac{d}{d x} ((4 x^{2} + 1)^{2})

\int \frac{1}{x ^{4} ( x ^{2} + 3 ) ^{\frac{1}{2}}} d x

x \to 0 + lim (e^{\frac{5}{x}})

\frac{d}{d x} ((x^{8} - 2)^{3})

What is the integral of sqrt(x+\sqrt{x^2+2)} ?
The integral of sqrt(x+\sqrt{x^2+2)} is 1/3 (x+sqrt(x^2+2))^{3/2}-2/(sqrt(x+\sqrt{x^2+2))}+C