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

The integral of sqrt(16+y^2) is 16(1/32 ysqrt(16+y^2)+1/2 ln((|y+sqrt(16+y^2)|)/4))+C

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integral of sqrt(16+y^2)

\int 16 + y^{2} d y

16 \frac{1}{32} y 16 + y^{2} + \frac{1}{2} ln \frac{y + 16 + y ^{2}}{4} + C

Solution steps

\int 16 + y^{2} d y

Apply Trigonometric Substitution

= \int 16 sec^{3} (u) d u

Take the constant out:

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

= 16 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 16 (\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) ∣

= 16 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{4} y)

= 16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} y ) ) sin ( arctan ( \frac{1}{4} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} y)) + sec (arctan (\frac{1}{4} y)))

Simplify

16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} y ) ) sin ( arctan ( \frac{1}{4} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} y)) + sec (arctan (\frac{1}{4} y))) : 16 \frac{1}{32} y 16 + y^{2} + \frac{1}{2} ln \frac{y + 16 + y ^{2}}{4}

= 16 \frac{1}{32} y 16 + y^{2} + \frac{1}{2} ln \frac{y + 16 + y ^{2}}{4}

Add a constant to the solution

= 16 \frac{1}{32} y 16 + y^{2} + \frac{1}{2} ln \frac{y + 16 + y ^{2}}{4} + C

What is the integral of sqrt(16+y^2) ?
The integral of sqrt(16+y^2) is 16(1/32 ysqrt(16+y^2)+1/2 ln((|y+sqrt(16+y^2)|)/4))+C