What is the integral of sqrt(5+w^2) ?

The integral of sqrt(5+w^2) is 5(1/10 wsqrt(5+w^2)+1/2 ln| 1/(sqrt(5))w+sqrt(1/5 (5+w^2))|)+C

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integral of sqrt(5+w^2)

\int 5 + w^{2} d w

5 (\frac{1}{10} w 5 + w^{2} + \frac{1}{2} ln \frac{1}{5} w + \frac{1}{5} (5 + w^{2})) + C

Solution steps

\int 5 + w^{2} d w

Apply Trigonometric Substitution

= \int 5 sec^{3} (u) d u

Take the constant out:

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

= 5 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 5 (\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) ∣

= 5 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{5} w)

= 5 \frac{sec ^{2} ( arctan ( \frac{1}{5} w ) ) sin ( arctan ( \frac{1}{5} w ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} w)) + sec (arctan (\frac{1}{5} w))

Simplify

5 \frac{sec ^{2} ( arctan ( \frac{1}{5} w ) ) sin ( arctan ( \frac{1}{5} w ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} w)) + sec (arctan (\frac{1}{5} w)) : 5 (\frac{1}{10} w 5 + w^{2} + \frac{1}{2} ln \frac{1}{5} w + \frac{1}{5} (5 + w^{2}))

= 5 (\frac{1}{10} w 5 + w^{2} + \frac{1}{2} ln \frac{1}{5} w + \frac{1}{5} (5 + w^{2}))

Add a constant to the solution

= 5 (\frac{1}{10} w 5 + w^{2} + \frac{1}{2} ln \frac{1}{5} w + \frac{1}{5} (5 + w^{2})) + C

What is the integral of sqrt(5+w^2) ?
The integral of sqrt(5+w^2) is 5(1/10 wsqrt(5+w^2)+1/2 ln| 1/(sqrt(5))w+sqrt(1/5 (5+w^2))|)+C