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

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

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

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

\frac{2 + ln ( 1 + 2 )}{4}

Decimal

0.57389 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} \frac{1}{2} sec^{3} (u) d u

= \int_{0}^{\frac{π}{4}} \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}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

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

Simplify

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

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