What is the integral of (x^2+4)^{1/2} ?

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

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integral of (x^2+4)^{1/2}

\int (x^{2} + 4)^{\frac{1}{2}} d x

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

Solution steps

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

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

Add a constant to the solution

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

\int_{0}^{y} 0.125 d x

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

n = 1 \sum \infty n^{2} (\frac{2}{3})^{n}

\frac{d}{d x} (7 x - 1)

\frac{d}{d x} (x (x - a) ln (x + 1 + b^{2}))

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