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

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

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integral of sqrt(4r^2+1)

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

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

Solution steps

\int 4 r^{2} + 1 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

\int_{- \frac{1}{2}}^{0} (\frac{1}{2} + 2 y + 3 z) d y

\frac{\partial}{\partial x} (e^{x s i n (y)})

1 + 2 x

x \to 2 lim (\frac{3 x + 4}{8 x - 1})

\int \frac{1}{u 3 - u ^{2}} d u

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