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

θ \to 0 lim (\frac{sin ( θ )}{sin ( 2 θ )})

\frac{\partial}{\partial x} (5 x cos (x) cos (y))

n = 1 \sum \infty 0. 7^{n}

\frac{d x}{d t} = 1 - x - x^{2}

e^{\frac{z}{z - 1}}

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