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

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

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integral of sqrt(4t^2+1)

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

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

Solution steps

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

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 t)

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (- ∣ x - 2 ∣ + 3)

\frac{\partial}{\partial x} (\frac{x + y}{x \cdot y})

\frac{d}{d x} (lo g_{3} (4 x))

\frac{d}{d θ} (\frac{8}{2 + cot ( θ ) sin ( θ )})

\int \frac{1}{x ( ln ( x ) ) ^{4}} d x

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