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

The integral of sqrt(((x^2+4))) 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(((x^2+4)))

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

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

Solution steps

\int x^{2} + 4 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 - 5 lim (12 x + 1)

h (x) = (x + x^{- 1})^{3}

\int e^{4 x + 9} d x

\frac{d}{d x} (4 x (x^{2} - 2))

f (x) = sin (cos (7 x))

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