What is the integral of sqrt(16+4x^2) ?

The integral of sqrt(16+4x^2) is xsqrt(x^2+4)+4ln(1/2 |x+sqrt(x^2+4)|)+C

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integral of sqrt(16+4x^2)

\int 16 + 4 x^{2} d x

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

Solution steps

\int 16 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int 8 sec^{3} (u) d u

= \int 8 sec^{3} (u) d u

Take the constant out:

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

= 8 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 8 (\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) ∣

= 8 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{4} x)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sqrt(16+4x^2) ?
The integral of sqrt(16+4x^2) is xsqrt(x^2+4)+4ln(1/2 |x+sqrt(x^2+4)|)+C