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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

= \int 2 sec^{3} (u) d u

Take the constant out:

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

= 2 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 2 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 2 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

= 2 \frac{sec ^{2} ( arctan ( \frac{1}{2 ^{\frac{1}{2}}} x ) ) sin ( arctan ( \frac{1}{2 ^{\frac{1}{2}}} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{2 ^{\frac{1}{2}}} x))

Simplify

2 \frac{sec ^{2} ( arctan ( \frac{1}{2 ^{\frac{1}{2}}} x ) ) sin ( arctan ( \frac{1}{2 ^{\frac{1}{2}}} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{2 ^{\frac{1}{2}}} x)) : \frac{1}{2} x (x^{2} + 2)^{\frac{1}{2}} + ln \frac{1}{2 ^{\frac{1}{2}}} x + \frac{1}{2} (2 + x^{2})

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

Add a constant to the solution

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

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