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

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

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

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

\frac{1}{4} 2 x 4 x^{2} + y^{2} + y^{2} ln \frac{2 x + y ^{2} + 4 x ^{2}}{∣ y ∣} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{y ^{2} sec ^{3} ( u )}{2} d u

= \int \frac{y ^{2} sec ^{3} ( u )}{2} d u

Take the constant out:

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

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

Apply Integral Reduction

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

Substitute back

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

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

Simplify

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

= \frac{1}{4} 2 x 4 x^{2} + y^{2} + y^{2} ln \frac{2 x + y ^{2} + 4 x ^{2}}{∣ y ∣}

Add a constant to the solution

= \frac{1}{4} 2 x 4 x^{2} + y^{2} + y^{2} ln \frac{2 x + y ^{2} + 4 x ^{2}}{∣ y ∣} + C

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