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

The integral of sqrt(x^2+1/4) is 1/8 (2xsqrt(4x^2+1)+ln|2x+sqrt(1+4x^2)|)+C

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

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

\frac{1}{8} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ^{3} ( u )}{4} d u

= \int \frac{sec ^{3} ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{4} (\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}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{\frac{1}{2}} x)

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

Simplify

\frac{1}{4} \frac{sec ^{2} ( arctan ( \frac{1}{\frac{1}{2}} x ) ) sin ( arctan ( \frac{1}{\frac{1}{2}} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{\frac{1}{2}} x)) + sec (arctan (\frac{1}{\frac{1}{2}} x)) : \frac{1}{8} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2})

= \frac{1}{8} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2})

Add a constant to the solution

= \frac{1}{8} (2 x 4 x^{2} + 1 + ln 2 x + 1 + 4 x^{2}) + C

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What is the integral of sqrt(x^2+1/4) ?
The integral of sqrt(x^2+1/4) is 1/8 (2xsqrt(4x^2+1)+ln|2x+sqrt(1+4x^2)|)+C