What is the integral of sqrt(r^2+4) ?

The integral of sqrt(r^2+4) is 4(1/8 rsqrt(4+r^2)+1/2 ln((|r+sqrt(4+r^2)|)/2))+C

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integral of sqrt(r^2+4)

\int r^{2} + 4 d r

4 (\frac{1}{8} r 4 + r^{2} + \frac{1}{2} ln (\frac{r + 4 + r ^{2}}{2})) + C

Solution steps

\int r^{2} + 4 d r

Apply Trigonometric Substitution

: \int 4 sec^{3} (u) d u

= \int 4 sec^{3} (u) d u

Take the constant out:

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

= 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 4 (\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) ∣

= 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{2} r)

= 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} r ) ) sin ( arctan ( \frac{1}{2} r ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} r)) + sec (arctan (\frac{1}{2} r)))

Simplify

4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} r ) ) sin ( arctan ( \frac{1}{2} r ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} r)) + sec (arctan (\frac{1}{2} r))) : 4 (\frac{1}{8} r 4 + r^{2} + \frac{1}{2} ln (\frac{r + 4 + r ^{2}}{2}))

= 4 (\frac{1}{8} r 4 + r^{2} + \frac{1}{2} ln (\frac{r + 4 + r ^{2}}{2}))

Add a constant to the solution

= 4 (\frac{1}{8} r 4 + r^{2} + \frac{1}{2} ln (\frac{r + 4 + r ^{2}}{2})) + C

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What is the integral of sqrt(r^2+4) ?
The integral of sqrt(r^2+4) is 4(1/8 rsqrt(4+r^2)+1/2 ln((|r+sqrt(4+r^2)|)/2))+C