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

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

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integral of sqrt(1+4r^2)

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

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

Solution steps

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

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 r)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (1 + 7 x)

\int \frac{( 1 - z )}{z z} d z

\int - 2 x^{2} + 2 d x

7 sin^{2} (t)

cos (π x)

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