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

The integral of sqrt(4+y^2) is 4(1/8 ysqrt(4+y^2)+1/2 ln((|y+sqrt(4+y^2)|)/2))+C

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integral of sqrt(4+y^2)

\int 4 + y^{2} d y

4 \frac{1}{8} y 4 + y^{2} + \frac{1}{2} ln \frac{y + 4 + y ^{2}}{2} + C

Solution steps

\int 4 + y^{2} d y

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} y)

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

Simplify

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

= 4 \frac{1}{8} y 4 + y^{2} + \frac{1}{2} ln \frac{y + 4 + y ^{2}}{2}

Add a constant to the solution

= 4 \frac{1}{8} y 4 + y^{2} + \frac{1}{2} ln \frac{y + 4 + y ^{2}}{2} + C

\int \frac{14}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

f (x) = arcsin (4 x)

x \to \infty lim (arctan (x^{2} - x^{5}))

y = e^{x} ln (x), (1, 0)

\frac{\partial}{\partial y} (e^{x} y^{2})

What is the integral of sqrt(4+y^2) ?
The integral of sqrt(4+y^2) is 4(1/8 ysqrt(4+y^2)+1/2 ln((|y+sqrt(4+y^2)|)/2))+C