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

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

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integral of 2sqrt(1+4x^2)

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

x 4 x^{2} + 1 + \frac{1}{2} ln 2 x + 1 + 4 x^{2} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int 1 + 4 x^{2} d x

Apply Trigonometric Substitution

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

= 2 \cdot \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

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

Apply Integral Reduction

= 2 \cdot \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) ∣

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

Substitute back

u = arctan (2 x)

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

Simplify

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

= x 4 x^{2} + 1 + \frac{1}{2} ln 2 x + 1 + 4 x^{2}

Add a constant to the solution

= x 4 x^{2} + 1 + \frac{1}{2} ln 2 x + 1 + 4 x^{2} + C

\frac{d}{d x} (x^{0.4} y^{0.6})

k \cdot d t = (p \cdot m - p^{2})^{- 1} \cdot d p

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

y = 3 - x^{2}, y = 6 - 4 x

x \to 2 + lim (2 x - [x])

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