What is the integral of sqrt(1+4y^2) ?

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

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integral of sqrt(1+4y^2)

\int 1 + 4 y^{2} d y

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

Solution steps

\int 1 + 4 y^{2} d y

Apply Trigonometric Substitution

: \int \frac{1}{2} sec^{3} (u) d u

= \int \frac{1}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

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

Substitute back

u = arctan (2 y)

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

Simplify

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

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

Add a constant to the solution

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

\frac{ln ( x )}{x ^{3}}

y = \frac{5 x ^{2} + 3}{x ^{2} + 1}

\frac{\partial}{\partial x} (cos (\frac{x}{2}))

x \to 7 lim (2 x^{- 1})

f (x) = \frac{x ^{12}}{x ^{2}}

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