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

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

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integral of sqrt(1+4θ^2)

\int 1 + 4 θ^{2} d θ

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

Solution steps

\int 1 + 4 θ^{2} d θ

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 θ)

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

Simplify

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

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

Add a constant to the solution

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

e^{2 t} (t - 1)^{2}

\frac{d}{d x} ((1 - 3 x) arccoth (3 x))

y^{^{'}} + 4 x y^{3} = 0

\frac{d}{d x} (\frac{4 x ^{3}}{3} + cos (x) + 1)

\frac{d}{d x} (sin (n x))

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