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

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

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

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2} d u

= \frac{1}{2} \cdot \int \frac{sec ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{2} \cdot \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (2 x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d t} + \frac{2}{t} y = 4 t - 3 + \frac{2}{t}

x \to 10 lim (\frac{x - 1 - 3}{x - 10})

7 u + 2 u

n = 0 \sum \infty (sin (100))^{n}

\int \frac{x}{x} d x

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