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

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

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integral of sqrt(1+4t^2)

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

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

Solution steps

\int 1 + 4 t^{2} 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 2 + lim (\frac{- 5}{x - 2})

x \to - 4 + lim (\frac{3}{x ^{2} - 16})

f (x) = x^{3} - 6 x + 9

\int (\frac{1}{2 x}) d x

\frac{d}{d x} (\frac{2 x - 3}{2 x + 5})

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