What is the integral from 0 to 1 of sqrt(1+4x^2) ?

The integral from 0 to 1 of sqrt(1+4x^2) is 1/2 (sqrt(5)+1/2 ln(2+sqrt(5)))

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integral from 0 to 1 of sqrt(1+4x^2)

\int_{0}^{1} 1 + 4 x^{2} d x

\frac{1}{2} (5 + \frac{1}{2} ln (2 + 5))

Decimal Notation

1.47894 \dots

Solution steps

\int_{0}^{1} 1 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (2)} \frac{1}{2} sec^{3} (u) d u

= \int_{0}^{a r c t a n (2)} \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_{0}^{a r c t a n (2)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec (u) d u)

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Compute the boundaries:

5

= \frac{1}{2} (5 + \frac{1}{2} ln (2 + 5))

\frac{3 x - 2}{2 x + 1}

f (x) = e^{t a n (θ)}

\frac{d y}{d θ} = sin (θ)

\int x e^{x y} + 2 x y + \frac{1}{x}

x \to - 2 lim (\frac{( x - 2 )}{x + 2})

What is the integral from 0 to 1 of sqrt(1+4x^2) ?
The integral from 0 to 1 of sqrt(1+4x^2) is 1/2 (sqrt(5)+1/2 ln(2+sqrt(5)))