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

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

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

\int \frac{x ^{2} + 4}{2} d x

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

Solution steps

\int \frac{x ^{2} + 4}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{2} + 4 d x

Apply Trigonometric Substitution

= \frac{1}{2} \cdot \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

= \frac{1}{2} \cdot 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2} \cdot 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) ∣

= \frac{1}{2} \cdot 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{2} x)

= \frac{1}{2} \cdot 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

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

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

Add a constant to the solution

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

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