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

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

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integral of sqrt(4+x^2)

\int 4 + x^{2} d x

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

Solution steps

\int 4 + x^{2} d x

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} x)

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

Simplify

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

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

Add a constant to the solution

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

x \to 3 lim (\frac{1 - x}{x ^{2} - 9})

d y - (y - 9)^{2} \cdot d x = 0

\frac{d}{d x} (\frac{cos ( m x )}{sin ( n x )})

\int_{0}^{4} (9 e^{- 2 x}) d x

\int \frac{4}{100 - t} d t

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