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

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

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

\int 1 + (2 x)^{2} d x

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

Solution steps

\int 1 + (2 x)^{2} d x

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 x)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 2 y^{^{'}} + y = 12 e - t

\int csc^{2} (θ) - 1 d θ

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

\frac{d}{d x} ((4 x - 3)^{8})

t^{2} + 4

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