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

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

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integral of sqrt(x^2+y)

\int x^{2} + y d x

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

Solution steps

\int x^{2} + y d x

Apply Trigonometric Substitution

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

Take the constant out:

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

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

Apply Integral Reduction

= y (\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) ∣

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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